blocks|key|282307|text|没有明确的答案，因为多个键可以映射到相同的值。如果您使用自己的代码强制唯一性，最好的解决方案是创建一个使用两个Hashmap来跟踪双向映射的类。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|282308|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

There is no unambiguous answer, because multiple keys can map to the same value. If you are enforcing unique-ness with your own code, the best solution is to create a class that uses two Hashmaps to track the mappings in both directions.

blocks|key|33480|text|要查找映射到该值的所有键，请使用map.entrySet()遍历散列映射中的所有对。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|33481|entityMap^0|G|E|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

To find all the keys that map to that value, iterate through all the pairs in the hashmap, using <code>map.entrySet()</code>.

blocks|key|1437747|text|听起来最好的方法是使用map.entrySet()遍历条目，因为map.containsValue()可能无论如何都会这样做。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1437748|entityMap^0|B|E|W|J|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

It sounds like the best way is for you to iterate over entries using <code>map.entrySet()</code> since <code>map.containsValue()</code> probably does this anyway.

blocks|key|1584213|text|如果您选择使用Commons+Collections+library而不是标准的Java+Collections框架，您可以很容易地实现这一点。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1584214|集合库中的BidiMap接口是一个双向映射，允许您将键映射到值(如法线映射)，也可以将值映射到键，从而允许您在两个方向上执行查找。getKey()方法支持获取值的键。|style|CODE|1584215|但是有一个警告，双向映射不能将多个值映射到键，因此，除非您的数据集在键和值之间有1:1的映射，否则不能使用双向映射。|1584216|1584217|如果希望依赖Java+Collections+API，则在将值插入映射时，必须确保键和值之间的1:1关系。这说起来容易做起来难。|1584218|一旦您能够确保这一点，就可以使用entrySet()方法来获取映射中的一组条目(映射)。获得类型为Map.Entry的集合后，遍历条目，将stored+value与预期值进行比较，并获得corresponding+key。|1584219|1584220|可以在Google+Guava和重构的Apache库(后者不是Commons-Collections项目)中找到对带有泛型的双向映射的支持。感谢Esko指出Apache+Commons+Collections中缺少的通用支持。将集合与泛型结合使用可以提高代码的可维护性。|1584221|1584222|从version+4.0开始，官方的Apache+Commons+Collections™库就支持。|1584223|有关现在支持summary泛型的BidiMap、OrderedBidiMap和SortedBidiMap接口的可用实现的列表，请参阅“org.apache.commons.collection+tions4.bidimap”包的Java页面。|1584224|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/collections/|1|http://commons.apache.org/proper/commons-collections/apidocs/org/apache/commons/collections4/BidiMap.html|2|http://commons.apache.org/proper/commons-collections/apidocs/org/apache/commons/collections4/BidiMap.html#getKey%2528java.lang.Object%2529|3|https://docs.oracle.com/javase/7/docs/api/java/util/Map.html#entrySet()|4|https://docs.oracle.com/javase/7/docs/api/java/util/Map.Entry.html|5|https://docs.oracle.com/javase/7/docs/api/java/util/Map.Entry.html#getValue()|6|https://docs.oracle.com/javase/7/docs/api/java/util/Map.Entry.html#getKey())|7|https://github.com/google/guava|8|http://sourceforge.net/projects/collections/|9|https://commons.apache.org/proper/commons-collections/release_4_0.html|10|https://commons.apache.org/proper/commons-collections/javadocs/api-4.4/org/apache/commons/collections4/bidimap/package-summary.html|11|https://commons.apache.org/proper/commons-collections/javadocs/api-4.4/org/apache/commons/collections4/BidiMap.html|12|https://commons.apache.org/proper/commons-collections/javadocs/api-4.4/org/apache/commons/collections4/OrderedBidiMap.html|13|https://commons.apache.org/proper/commons-collections/javadocs/api-4.4/org/apache/commons/collections4/SortedBidiMap.html^0|7|R|0|0|5|7|1T|8|5|7|1|1T|8|2|0|0|0|0|G|A|1D|9|G|A|3|1D|9|4|1X|C|5|2L|H|6|0|0|3|C|7|V|J|8|0|0|1|B|9|0|G|7|O|E|13|D|6|7|A|G|7|B|O|E|C|13|D|D|0^^$0|@$1|2|3|4|5|6|7|1U|8|@]|9|@$A|1V|B|1W|1|1X]]|C|$]]|$1|D|3|E|5|6|7|1Y|8|@$A|1Z|B|20|F|G]|$A|21|B|22|F|G]]|9|@$A|23|B|24|1|25]|$A|26|B|27|1|28]]|C|$]]|$1|H|3|I|5|6|7|29|8|@]|9|@]|C|$]]|$1|J|3|-4|5|6|7|2A|8|@]|9|@]|C|$]]|$1|K|3|L|5|6|7|2B|8|@]|9|@]|C|$]]|$1|M|3|N|5|6|7|2C|8|@$A|2D|B|2E|F|G]|$A|2F|B|2G|F|G]]|9|@$A|2H|B|2I|1|2J]|$A|2K|B|2L|1|2M]|$A|2N|B|2O|1|2P]|$A|2Q|B|2R|1|2S]]|C|$]]|$1|O|3|-4|5|6|7|2T|8|@]|9|@]|C|$]]|$1|P|3|Q|5|6|7|2U|8|@]|9|@$A|2V|B|2W|1|2X]|$A|2Y|B|2Z|1|30]]|C|$]]|$1|R|3|-4|5|6|7|31|8|@]|9|@]|C|$]]|$1|S|3|T|5|6|7|32|8|@]|9|@$A|33|B|34|1|35]]|C|$]]|$1|U|3|V|5|6|7|36|8|@$A|37|B|38|F|G]|$A|39|B|3A|F|G]|$A|3B|B|3C|F|G]]|9|@$A|3D|B|3E|1|3F]|$A|3G|B|3H|1|3I]|$A|3J|B|3K|1|3L]|$A|3M|B|3N|1|3O]]|C|$]]|$1|W|3|-4|5|6|7|3P|8|@]|9|@]|C|$]]]|X|$Y|$5|Z|10|11|C|$12|13]]|14|$5|Z|10|11|C|$12|15]]|16|$5|Z|10|11|C|$12|17]]|18|$5|Z|10|11|C|$12|19]]|1A|$5|Z|10|11|C|$12|1B]]|1C|$5|Z|10|11|C|$12|1D]]|1E|$5|Z|10|11|C|$12|1F]]|1G|$5|Z|10|11|C|$12|1H]]|1I|$5|Z|10|11|C|$12|1J]]|1K|$5|Z|10|11|C|$12|1L]]|1M|$5|Z|10|11|C|$12|1N]]|1O|$5|Z|10|11|C|$12|1P]]|1Q|$5|Z|10|11|C|$12|1R]]|1S|$5|Z|10|11|C|$12|1T]]]]

If you choose to use the <a href="http://commons.apache.org/collections/" rel="noreferrer">Commons Collections library</a> instead of the standard Java Collections framework, you can achieve this with ease.
The <a href="http://commons.apache.org/proper/commons-collections/apidocs/org/apache/commons/collections4/BidiMap.html" rel="noreferrer"><code>BidiMap</code></a> interface in the Collections library is a bi-directional map, allowing you to map a key to a value (like normal maps), and also to map a value to a key, thus allowing you to perform lookups in both directions. Obtaining a key for a value is supported by the <a href="http://commons.apache.org/proper/commons-collections/apidocs/org/apache/commons/collections4/BidiMap.html#getKey%28java.lang.Object%29" rel="noreferrer"><code>getKey()</code></a> method.
There is a caveat though, bidi maps cannot have multiple values mapped to keys, and hence unless your data set has 1:1 mappings between keys and values, you cannot use bidi maps.
<hr />
If you want to rely on the Java Collections API, you will have to ensure the 1:1 relationship between keys and values at the time of inserting the value into the map. This is easier said than done.
Once you can ensure that, use the <a href="https://docs.oracle.com/javase/7/docs/api/java/util/Map.html#entrySet()" rel="noreferrer"><code>entrySet()</code></a> method to obtain the set of entries (mappings) in the Map. Once you have obtained the set whose type is <a href="https://docs.oracle.com/javase/7/docs/api/java/util/Map.Entry.html" rel="noreferrer"><code>Map.Entry</code></a>, iterate through the entries, comparing the <a href="https://docs.oracle.com/javase/7/docs/api/java/util/Map.Entry.html#getValue()" rel="noreferrer">stored value</a> against the expected, and obtain the <a href="https://docs.oracle.com/javase/7/docs/api/java/util/Map.Entry.html#getKey()" rel="noreferrer">corresponding key</a>.
<hr />
Support for bidi maps with generics can be found in <a href="https://github.com/google/guava" rel="noreferrer">Google Guava</a> and the refactored <a href="http://sourceforge.net/projects/collections/" rel="noreferrer">Commons-Collections</a> libraries (the latter is not an Apache project). Thanks to Esko for pointing out the missing generic support in Apache Commons Collections. Using collections with generics makes more maintainable code.
<hr />
Since <a href="https://commons.apache.org/proper/commons-collections/release_4_0.html" rel="noreferrer">version 4.0</a> the official Apache Commons Collections™ library supports <a href="https://en.wikipedia.org/wiki/Generics_in_Java" rel="noreferrer">generics</a>.
See the <a href="https://commons.apache.org/proper/commons-collections/javadocs/api-4.4/org/apache/commons/collections4/bidimap/package-summary.html" rel="noreferrer">summary</a> page of the &quot;org.apache.commons.collections4.bidimap&quot; package for the list of available implementations of the <a href="https://commons.apache.org/proper/commons-collections/javadocs/api-4.4/org/apache/commons/collections4/BidiMap.html" rel="noreferrer"><code>BidiMap</code></a>, <a href="https://commons.apache.org/proper/commons-collections/javadocs/api-4.4/org/apache/commons/collections4/OrderedBidiMap.html" rel="noreferrer"><code>OrderedBidiMap</code></a> and <a href="https://commons.apache.org/proper/commons-collections/javadocs/api-4.4/org/apache/commons/collections4/SortedBidiMap.html" rel="noreferrer"><code>SortedBidiMap</code></a> interfaces that now support Java generics.

blocks|key|1437854|text|我认为你的选择是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1437855|1437856|使用了为此而构建的地图实现，比如来自谷歌集合的BiMap。注意，谷歌集合BiMap要求值和键的唯一性，但它在两个方向上都提供了高性能，并维护了两个映射-一个是key+->+value，另一个是value+->+key|unordered-list-item|offset|length|1437857|遍历entrySet()并找到与该值匹配的键。这是最慢的方法，因为它需要遍历整个集合，而其他两个方法不需要这样做。|style|CODE|1437858|1437859|entityMap|0|LINK|mutability|MUTABLE|url|http://google-collections.googlecode.com/svn/trunk/javadoc/index.html?com/google/common/collect/BiMap.html^0|0|0|N|5|0|0|2|A|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|V|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|W|8|@]|9|@$F|X|G|Y|1|Z]]|A|$]]|$1|H|3|I|5|E|7|10|8|@$F|11|G|12|J|K]]|9|@]|A|$]]|$1|L|3|-4|5|6|7|13|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

I think your choices are

<ul>
<li>Use a map implementation built for this, like the <a href="http://google-collections.googlecode.com/svn/trunk/javadoc/index.html?com/google/common/collect/BiMap.html" rel="noreferrer">BiMap</a> from google collections. Note that the google collections BiMap requires uniqueless of values, as well as keys, but it provides high performance in both directions performance</li>
<li>Manually maintain two maps - one for key -> value, and another map for value -> key</li>
<li>Iterate through the <code>entrySet()</code> and to find the keys which match the value. This is the slowest method, since it requires iterating through the entire collection, while the other two methods don't require that.</li>
</ul>

blocks|key|282448|text|如果您在自己的代码中构建映射，请尝试将键和值放在映射中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|282449|public+class+KeyValue+{
++++public+Object+key;
++++public+Object+value;
++++public+KeyValue(Object+key,+Object+value)+{+...+}
}

map.put(key,+new+KeyValue(key,+value));|code-block|syntax|javascript|282450|那么当你有一个值的时候，你也就有了这个键。|282451|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

If you build the map in your own code, try putting the key and value in the map together:

<pre><code>public class KeyValue {
 public Object key;
 public Object value;
 public KeyValue(Object key, Object value) { ... }
}

map.put(key, new KeyValue(key, value));
</code></pre>

Then when you have a value, you also have the key.

blocks|key|1584300|text|是的，您必须遍历hashmap，除非您实现了这些不同答案所暗示的内容。我只需获取keySet()，遍历该集合，并保留(第一个)获得匹配值的(第一个)键，而不是摆弄entrySet。如果您需要与该值匹配的所有键，显然您必须执行所有操作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1584301|正如乔纳斯所建议的那样，这可能已经是containsValue方法正在做的事情了，所以你可以跳过所有的测试，并且每次都进行迭代(或者编译器可能已经消除了冗余，谁知道呢)。|1584302|另外，相对于其他答案，如果您的反向映射看起来像|1584303|Map<Value,+Set<Key>>|code-block|syntax|javascript|1584304|如果您需要该功能，您可以处理非唯一键->值映射(将它们解开)。这将把fine整合到人们在这里使用两个地图建议的任何解决方案中。|1584305|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Q|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|R|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Yes, you have to loop through the hashmap, unless you implement something along the lines of what these various answers suggest. Rather than fiddling with the entrySet, I'd just get the keySet(), iterate over that set, and keep the (first) key that gets you your matching value. If you need all the keys that match that value, obviously you have to do the whole thing.

As Jonas suggests, this might already be what the containsValue method is doing, so you might just skip that test all-together, and just do the iteration every time (or maybe the compiler will already eliminate the redundancy, who knows).

Also, relative to the other answers, if your reverse map looks like

<pre><code>Map&lt;Value, Set&lt;Key&gt;&gt;
</code></pre>

you can deal with non-unique key->value mappings, if you need that capability (untangling them aside). That would incorporate fine into any of the solutions people suggest here using two maps.

blocks|key|1584314|text|如果你的数据结构在键和值之间有多对一的映射，你应该遍历条目并选择所有合适的键：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1584315|public+static+<T,+E>+Set<T>+getKeysByValue(Map<T,+E>+map,+E+value)+{
++++Set<T>+keys+=+new+HashSet<T>();
++++for+(Entry<T,+E>+entry+:+map.entrySet())+{
++++++++if+(Objects.equals(value,+entry.getValue()))+{
++++++++++++keys.add(entry.getKey());
++++++++}
++++}
++++return+keys;
}|code-block|syntax|javascript|1584316|如果是一对一的关系，可以返回第一个匹配的key：|1584317|public+static+<T,+E>+T+getKeyByValue(Map<T,+E>+map,+E+value)+{
++++for+(Entry<T,+E>+entry+:+map.entrySet())+{
++++++++if+(Objects.equals(value,+entry.getValue()))+{
++++++++++++return+entry.getKey();
++++++++}
++++}
++++return+null;
}|1584318|在Java+8中：|1584319|public+static+<T,+E>+Set<T>+getKeysByValue(Map<T,+E>+map,+E+value)+{
++++return+map.entrySet()
++++++++++++++.stream()
++++++++++++++.filter(entry+->+Objects.equals(entry.getValue(),+value))
++++++++++++++.map(Map.Entry::getKey)
++++++++++++++.collect(Collectors.toSet());
}|1584320|此外，对于Guava用户，BiMap可能很有用。例如：|offset|length|1584321|BiMap<Token,+Character>+tokenToChar+=+
++++ImmutableBiMap.of(Token.LEFT_BRACKET,+'[',+Token.LEFT_PARENTHESIS,+'(');
Token+token+=+tokenToChar.inverse().get('(');
Character+c+=+tokenToChar.get(token);|1584322|entityMap|0|LINK|mutability|MUTABLE|url|https://google.github.io/guava/releases/snapshot/api/docs/com/google/common/collect/BiMap.html^0|0|0|0|0|0|0|D|5|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|14|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|16|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|18|8|@]|9|@$Q|19|R|1A|1|1B]]|A|$]]|$1|S|3|T|5|D|7|1C|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]]]

If your data structure has many-to-one mapping between keys and values you should iterate over entries and pick all suitable keys:

<pre><code>public static &lt;T, E&gt; Set&lt;T&gt; getKeysByValue(Map&lt;T, E&gt; map, E value) {
 Set&lt;T&gt; keys = new HashSet&lt;T&gt;();
 for (Entry&lt;T, E&gt; entry : map.entrySet()) {
 if (Objects.equals(value, entry.getValue())) {
 keys.add(entry.getKey());
 }
 }
 return keys;
}
</code></pre>

In case of one-to-one relationship, you can return the first matched key:

<pre><code>public static &lt;T, E&gt; T getKeyByValue(Map&lt;T, E&gt; map, E value) {
 for (Entry&lt;T, E&gt; entry : map.entrySet()) {
 if (Objects.equals(value, entry.getValue())) {
 return entry.getKey();
 }
 }
 return null;
}
</code></pre>

In Java 8:

<pre><code>public static &lt;T, E&gt; Set&lt;T&gt; getKeysByValue(Map&lt;T, E&gt; map, E value) {
 return map.entrySet()
 .stream()
 .filter(entry -&gt; Objects.equals(entry.getValue(), value))
 .map(Map.Entry::getKey)
 .collect(Collectors.toSet());
}
</code></pre>

Also, for Guava users, <a href="https://google.github.io/guava/releases/snapshot/api/docs/com/google/common/collect/BiMap.html" rel="noreferrer">BiMap</a> may be useful. For example:

<pre><code>BiMap&lt;Token, Character&gt; tokenToChar = 
 ImmutableBiMap.of(Token.LEFT_BRACKET, '[', Token.LEFT_PARENTHESIS, '(');
Token token = tokenToChar.inverse().get('(');
Character c = tokenToChar.get(token);
</code></pre>

blocks|key|282506|text|恐怕你只能迭代你的地图了。我能想到的最短的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|282507|Iterator<Map.Entry<String,String>>+iter+=+map.entrySet().iterator();
while+(iter.hasNext())+{
++++Map.Entry<String,String>+entry+=+iter.next();
++++if+(entry.getValue().equals(value_you_look_for))+{
++++++++String+key_you_look_for+=+entry.getKey();
++++}
}|code-block|syntax|javascript|282508|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I'm afraid you'll just have to iterate your map. Shortest I could come up with:

<pre><code>Iterator&lt;Map.Entry&lt;String,String&gt;&gt; iter = map.entrySet().iterator();
while (iter.hasNext()) {
 Map.Entry&lt;String,String&gt; entry = iter.next();
 if (entry.getValue().equals(value_you_look_for)) {
 String key_you_look_for = entry.getKey();
 }
}
</code></pre>

blocks|key|33792|text|您可以使用以下代码使用值来获取密钥。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|33793|ArrayList+valuesList+=+new+ArrayList();
Set+keySet+=+initalMap.keySet();
ArrayList+keyList+=+new+ArrayList(keySet);

for(int+i+=+0+;+i+<+keyList.size()+;+i%2B%2B+)+{
++++valuesList.add(initalMap.get(keyList.get(i)));
}

Collections.sort(valuesList);
Map+finalMap+=+new+TreeMap();
for(int+i+=+0+;+i+<+valuesList.size()+;+i%2B%2B+)+{
++++String+value+=+(String)+valuesList.get(i);

++++for(+int+j+=+0+;+j+<+keyList.size()+;+j%2B%2B+)+{
++++++++if(initalMap.get(keyList.get(j)).equals(value))+{
++++++++++++finalMap.put(keyList.get(j),value);
++++++++}+++
++++}
}
System.out.println("fianl+map+---------------------->++"+%2B+finalMap);|code-block|syntax|javascript|33794|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can get the key using values using following code..

<pre><code>ArrayList valuesList = new ArrayList();
Set keySet = initalMap.keySet();
ArrayList keyList = new ArrayList(keySet);

for(int i = 0 ; i &lt; keyList.size() ; i++ ) {
 valuesList.add(initalMap.get(keyList.get(i)));
}

Collections.sort(valuesList);
Map finalMap = new TreeMap();
for(int i = 0 ; i &lt; valuesList.size() ; i++ ) {
 String value = (String) valuesList.get(i);

 for( int j = 0 ; j &lt; keyList.size() ; j++ ) {
 if(initalMap.get(keyList.get(j)).equals(value)) {
 finalMap.put(keyList.get(j),value);
 } 
 }
}
System.out.println("fianl map ----------------------&gt; " + finalMap);
</code></pre>

blocks|key|1584368|text|public+class+NewClass1+{

++++public+static+void+main(String[]+args)+{
+++++++Map<Integer,+String>+testMap+=+new+HashMap<Integer,+String>();
++++++++testMap.put(10,+"a");
++++++++testMap.put(20,+"b");
++++++++testMap.put(30,+"c");
++++++++testMap.put(40,+"d");
++++++++for+(Entry<Integer,+String>+entry+:+testMap.entrySet())+{
++++++++++++if+(entry.getValue().equals("c"))+{
++++++++++++++++System.out.println(entry.getKey());
++++++++++++}
++++++++}
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1584369|一些附加信息...可能对你有用，|unstyled|offset|length|style|BOLD|1584370|如果你的hashmap真的很大，上面的方法可能不是很好。如果hashmap包含唯一键到唯一值映射，则可以再维护一个包含值到键映射的hashmap。|1584371|也就是说，您必须维护两个hashmap|1584372|1.+Key+to+value

2.+Value+to+key+|1584373|在这种情况下，您可以使用第二个hashmap来获取键。|1584374|entityMap^0|0|0|G|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|V|8|@$G|W|H|X|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$B|C]]|$1|Q|3|R|5|F|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|F|7|12|8|@]|9|@]|A|$]]]|T|$]]

<pre><code>public class NewClass1 {

 public static void main(String[] args) {
 Map&lt;Integer, String&gt; testMap = new HashMap&lt;Integer, String&gt;();
 testMap.put(10, "a");
 testMap.put(20, "b");
 testMap.put(30, "c");
 testMap.put(40, "d");
 for (Entry&lt;Integer, String&gt; entry : testMap.entrySet()) {
 if (entry.getValue().equals("c")) {
 System.out.println(entry.getKey());
 }
 }
 }
}
</code></pre>

Some additional info... May be useful to you

Above method may not be good if your hashmap is really big. If your hashmap contain unique key to unique value mapping, you can maintain one more hashmap that contain mapping from Value to Key.

That is you have to maintain two hashmaps

<pre><code>1. Key to value

2. Value to key 
</code></pre>

In that case you can use second hashmap to get key.

blocks|key|1581893|text|使用瘦包装器：HMap|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1581894|import+java.util.Collections;
import+java.util.HashMap;
import+java.util.Map;

public+class+HMap<K,+V>+{

+++private+final+Map<K,+Map<K,+V>>+map;

+++public+HMap()+{
++++++map+=+new+HashMap<K,+Map<K,+V>>();
+++}

+++public+HMap(final+int+initialCapacity)+{
++++++map+=+new+HashMap<K,+Map<K,+V>>(initialCapacity);
+++}

+++public+boolean+containsKey(final+Object+key)+{
++++++return+map.containsKey(key);
+++}

+++public+V+get(final+Object+key)+{
++++++final+Map<K,+V>+entry+=+map.get(key);
++++++if+(entry+!=+null)
+++++++++return+entry.values().iterator().next();
++++++return+null;
+++}

+++public+K+getKey(final+Object+key)+{
++++++final+Map<K,+V>+entry+=+map.get(key);
++++++if+(entry+!=+null)
+++++++++return+entry.keySet().iterator().next();
++++++return+null;
+++}

+++public+V+put(final+K+key,+final+V+value)+{
++++++final+Map<K,+V>+entry+=+map
++++++++++++.put(key,+Collections.singletonMap(key,+value));
++++++if+(entry+!=+null)
+++++++++return+entry.values().iterator().next();
++++++return+null;
+++}
}|code-block|syntax|javascript|1581895|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/jayenashar/Java-Search-library/blob/master/src/au/edu/unsw/cse/jayen/util/HMap.java^0|7|4|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Use a thin wrapper: <a href="https://github.com/jayenashar/Java-Search-library/blob/master/src/au/edu/unsw/cse/jayen/util/HMap.java" rel="nofollow">HMap</a>

<pre><code>import java.util.Collections;
import java.util.HashMap;
import java.util.Map;

public class HMap&lt;K, V&gt; {

 private final Map&lt;K, Map&lt;K, V&gt;&gt; map;

 public HMap() {
 map = new HashMap&lt;K, Map&lt;K, V&gt;&gt;();
 }

 public HMap(final int initialCapacity) {
 map = new HashMap&lt;K, Map&lt;K, V&gt;&gt;(initialCapacity);
 }

 public boolean containsKey(final Object key) {
 return map.containsKey(key);
 }

 public V get(final Object key) {
 final Map&lt;K, V&gt; entry = map.get(key);
 if (entry != null)
 return entry.values().iterator().next();
 return null;
 }

 public K getKey(final Object key) {
 final Map&lt;K, V&gt; entry = map.get(key);
 if (entry != null)
 return entry.keySet().iterator().next();
 return null;
 }

 public V put(final K key, final V value) {
 final Map&lt;K, V&gt; entry = map
 .put(key, Collections.singletonMap(key, value));
 if (entry != null)
 return entry.values().iterator().next();
 return null;
 }
}
</code></pre>

blocks|key|282562|text|import+java.util.ArrayList;
import+java.util.HashMap;
import+java.util.Iterator;
import+java.util.List;
import+java.util.Set;

public+class+M{
public+static+void+main(String[]+args)+{

++++++++HashMap<String,+List<String>>+resultHashMap+=+new+HashMap<String,+List<String>>();

++++++++Set<String>+newKeyList+=+resultHashMap.keySet();


++++++++for+(Iterator<String>+iterator+=+originalHashMap.keySet().iterator();+iterator.hasNext();)+{
++++++++++++String+hashKey+=+(String)+iterator.next();

++++++++++++if+(!newKeyList.contains(originalHashMap.get(hashKey)))+{
++++++++++++++++List<String>+loArrayList+=+new+ArrayList<String>();
++++++++++++++++loArrayList.add(hashKey);
++++++++++++++++resultHashMap.put(originalHashMap.get(hashKey),+loArrayList);
++++++++++++}+else+{
++++++++++++++++List<String>+loArrayList+=+resultHashMap.get(originalHashMap
++++++++++++++++++++++++.get(hashKey));
++++++++++++++++loArrayList.add(hashKey);
++++++++++++++++resultHashMap.put(originalHashMap.get(hashKey),+loArrayList);
++++++++++++}
++++++++}

++++++++System.out.println("Original+HashMap+:+"+%2B+originalHashMap);
++++++++System.out.println("Result+HashMap+:+"+%2B+resultHashMap);
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|282563|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.Set;

public class M{
public static void main(String[] args) {

 HashMap&lt;String, List&lt;String&gt;&gt; resultHashMap = new HashMap&lt;String, List&lt;String&gt;&gt;();

 Set&lt;String&gt; newKeyList = resultHashMap.keySet();


 for (Iterator&lt;String&gt; iterator = originalHashMap.keySet().iterator(); iterator.hasNext();) {
 String hashKey = (String) iterator.next();

 if (!newKeyList.contains(originalHashMap.get(hashKey))) {
 List&lt;String&gt; loArrayList = new ArrayList&lt;String&gt;();
 loArrayList.add(hashKey);
 resultHashMap.put(originalHashMap.get(hashKey), loArrayList);
 } else {
 List&lt;String&gt; loArrayList = resultHashMap.get(originalHashMap
 .get(hashKey));
 loArrayList.add(hashKey);
 resultHashMap.put(originalHashMap.get(hashKey), loArrayList);
 }
 }

 System.out.println("Original HashMap : " + originalHashMap);
 System.out.println("Result HashMap : " + resultHashMap);
 }
}
</code></pre>

blocks|key|1584392|text|public+static+class+SmartHashMap+<T1+extends+Object,+T2+extends+Object>+{
++++public+HashMap<T1,+T2>+keyValue;
++++public+HashMap<T2,+T1>+valueKey;

++++public+SmartHashMap(){
++++++++this.keyValue+=+new+HashMap<T1,+T2>();
++++++++this.valueKey+=+new+HashMap<T2,+T1>();
++++}

++++public+void+add(T1+key,+T2+value){
++++++++this.keyValue.put(key,+value);
++++++++this.valueKey.put(value,+key);
++++}

++++public+T2+getValue(T1+key){
++++++++return+this.keyValue.get(key);
++++}

++++public+T1+getKey(T2+value){
++++++++return+this.valueKey.get(value);
++++}

}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1584393|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public static class SmartHashMap &lt;T1 extends Object, T2 extends Object&gt; {
 public HashMap&lt;T1, T2&gt; keyValue;
 public HashMap&lt;T2, T1&gt; valueKey;

 public SmartHashMap(){
 this.keyValue = new HashMap&lt;T1, T2&gt;();
 this.valueKey = new HashMap&lt;T2, T1&gt;();
 }

 public void add(T1 key, T2 value){
 this.keyValue.put(key, value);
 this.valueKey.put(value, key);
 }

 public T2 getValue(T1 key){
 return this.keyValue.get(key);
 }

 public T1 getKey(T2 value){
 return this.valueKey.get(value);
 }

}
</code></pre>

blocks|key|33886|text|import+java.util.HashMap;
import+java.util.HashSet;
import+java.util.Set;

public+class+ValueKeysMap<K,+V>+extends+HashMap+<K,V>{
++++HashMap<V,+Set<K>>+ValueKeysMap+=+new+HashMap<V,+Set<K>>();

++++@Override
++++public+boolean+containsValue(Object+value)+{
++++++++return+ValueKeysMap.containsKey(value);
++++}

++++@Override
++++public+V+put(K+key,+V+value)+{
++++++++if+(containsValue(value))+{
++++++++++++Set<K>+keys+=+ValueKeysMap.get(value);
++++++++++++keys.add(key);
++++++++}+else+{
++++++++++++Set<K>+keys+=+new+HashSet<K>();
++++++++++++keys.add(key);
++++++++++++ValueKeysMap.put(value,+keys);
++++++++}
++++++++return+super.put(key,+value);
++++}

++++@Override
++++public+V+remove(Object+key)+{
++++++++V+value+=+super.remove(key);
++++++++Set<K>+keys+=+ValueKeysMap.get(value);
++++++++keys.remove(key);
++++++++if(keys.size()+==+0)+{
+++++++++++ValueKeysMap.remove(value);
++++++++}
++++++++return+value;
++++}

++++public+Set<K>+getKeys4ThisValue(V+value){
++++++++Set<K>+keys+=+ValueKeysMap.get(value);
++++++++return+keys;
++++}

++++public+boolean+valueContainsThisKey(K+key,+V+value){
++++++++if+(containsValue(value))+{
++++++++++++Set<K>+keys+=+ValueKeysMap.get(value);
++++++++++++return+keys.contains(key);
++++++++}
++++++++return+false;
++++}

++++/*
+++++*+Take+care+of+argument+constructor+and+other+api's+like+putAll
+++++*/
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|33887|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>import java.util.HashMap;
import java.util.HashSet;
import java.util.Set;

public class ValueKeysMap&lt;K, V&gt; extends HashMap &lt;K,V&gt;{
 HashMap&lt;V, Set&lt;K&gt;&gt; ValueKeysMap = new HashMap&lt;V, Set&lt;K&gt;&gt;();

 @Override
 public boolean containsValue(Object value) {
 return ValueKeysMap.containsKey(value);
 }

 @Override
 public V put(K key, V value) {
 if (containsValue(value)) {
 Set&lt;K&gt; keys = ValueKeysMap.get(value);
 keys.add(key);
 } else {
 Set&lt;K&gt; keys = new HashSet&lt;K&gt;();
 keys.add(key);
 ValueKeysMap.put(value, keys);
 }
 return super.put(key, value);
 }

 @Override
 public V remove(Object key) {
 V value = super.remove(key);
 Set&lt;K&gt; keys = ValueKeysMap.get(value);
 keys.remove(key);
 if(keys.size() == 0) {
 ValueKeysMap.remove(value);
 }
 return value;
 }

 public Set&lt;K&gt; getKeys4ThisValue(V value){
 Set&lt;K&gt; keys = ValueKeysMap.get(value);
 return keys;
 }

 public boolean valueContainsThisKey(K key, V value){
 if (containsValue(value)) {
 Set&lt;K&gt; keys = ValueKeysMap.get(value);
 return keys.contains(key);
 }
 return false;
 }

 /*
 * Take care of argument constructor and other api's like putAll
 */
}
</code></pre>

blocks|key|282581|text|您可以将键、值对及其倒数都插入到映射结构中|type|unstyled|depth|inlineStyleRanges|entityRanges|data|282582|map.put("theKey",+"theValue");
map.put("theValue",+"theKey");|code-block|syntax|javascript|282583|使用map.get("theValue")将返回"theKey“。|282584|这是一种快速而肮脏的方法，我制作了常量映射，它只适用于选定的几个数据集：|282585|282586|仅包含1到1对|unordered-list-item|282587|值集合与密钥集合不相交(1->2，2->3破坏它)|282588|282589|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|V|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|W|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|X|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|Y|8|@]|9|@]|A|$]]|$1|O|3|P|5|N|7|Z|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|S|$]]

You could insert both the key,value pair and its inverse into your map structure

<pre><code>map.put("theKey", "theValue");
map.put("theValue", "theKey");
</code></pre>

Using map.get("theValue") will then return "theKey".

It's a quick and dirty way that I've made constant maps, which will only work for a select few datasets:

<ul>
<li>Contains only 1 to 1 pairs</li>
<li>Set of values is disjoint from the set of keys (1->2, 2->3 breaks it)</li>
</ul>

blocks|key|33932|text|Iterator<Map.Entry<String,String>>+iterator+=+map.entrySet().iterator();
while+(iterator.hasNext())+{
++++Map.Entry<String,String>+entry+=+iterator.next();
++++if+(entry.getValue().equals(value_you_look_for))+{
++++++++String+key_you_look_for+=+entry.getKey();
}
}+|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|33933|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>Iterator&lt;Map.Entry&lt;String,String&gt;&gt; iterator = map.entrySet().iterator();
while (iterator.hasNext()) {
 Map.Entry&lt;String,String&gt; entry = iterator.next();
 if (entry.getValue().equals(value_you_look_for)) {
 String key_you_look_for = entry.getKey();
}
} 
</code></pre>

blocks|key|1584421|text|在java8中|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1584422|map.entrySet().stream().filter(entry+->+entry.getValue().equals(value))
++++.forEach(entry+->+System.out.println(entry.getKey()));|code-block|syntax|javascript|1584423|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In java8

<pre><code>map.entrySet().stream().filter(entry -&gt; entry.getValue().equals(value))
 .forEach(entry -&gt; System.out.println(entry.getKey()));
</code></pre>

blocks|key|1438274|text|我认为这是最好的解决方案，原始地址：Java2s|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1438275|++++import+java.util.HashMap;
++++import+java.util.Map;

++++++++public+class+Main+{

++++++++++public+static+void+main(String[]+argv)+{
++++++++++++Map<String,+String>+map+=+new+HashMap<String,+String>();
++++++++++++map.put("1","one");
++++++++++++map.put("2","two");
++++++++++++map.put("3","three");
++++++++++++map.put("4","four");

++++++++++++System.out.println(getKeyFromValue(map,"three"));
++++++++++}


//+hm+is+the+map+you+are+trying+to+get+value+from+it
++++++++++public+static+Object+getKeyFromValue(Map+hm,+Object+value)+{
++++++++++++for+(Object+o+:+hm.keySet())+{
++++++++++++++if+(hm.get(o).equals(value))+{
++++++++++++++++return+o;
++++++++++++++}
++++++++++++}
++++++++++++return+null;
++++++++++}
++++++++}|code-block|syntax|javascript|1438276|一个简单的用法:如果你把所有的数据都放在hasMap中，并且有item+=+"Automobile"，那么你就可以在hashMap中查找它的键。这是一个很好的解决方案。|1438277|getKeyFromValue(hashMap,+item);
System.out.println("getKeyFromValue(hashMap,+item):+"%2BgetKeyFromValue(hashMap,+item));|1438278|entityMap|0|LINK|mutability|MUTABLE|url|http://www.java2s.com/Code/Java/Collections-Data-Structure/GetakeyfromvaluewithanHashMap.htm^0|I|6|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

I think this is best solution, original address: <a href="http://www.java2s.com/Code/Java/Collections-Data-Structure/GetakeyfromvaluewithanHashMap.htm" rel="noreferrer">Java2s</a>

<pre><code> import java.util.HashMap;
 import java.util.Map;

 public class Main {

 public static void main(String[] argv) {
 Map&lt;String, String&gt; map = new HashMap&lt;String, String&gt;();
 map.put("1","one");
 map.put("2","two");
 map.put("3","three");
 map.put("4","four");

 System.out.println(getKeyFromValue(map,"three"));
 }


// hm is the map you are trying to get value from it
 public static Object getKeyFromValue(Map hm, Object value) {
 for (Object o : hm.keySet()) {
 if (hm.get(o).equals(value)) {
 return o;
 }
 }
 return null;
 }
 }
</code></pre>

An easy usage: 
if you put all data in hasMap and you have item = "Automobile", so you are looking its key in hashMap. that is good solution. 

<pre><code>getKeyFromValue(hashMap, item);
System.out.println("getKeyFromValue(hashMap, item): "+getKeyFromValue(hashMap, item));
</code></pre>

blocks|key|33991|text|/**
+*+This+method+gets+the+Key+for+the+given+Value
+*+@param+paramName
+*+@return
+*/
private+String+getKeyForValueFromMap(String+paramName)+{
++++String+keyForValue+=+null;
++++if(paramName!=null))+{
++++++++Set<Entry<String,String>>+entrySet+=+myMap().entrySet();
++++++++if(entrySet!=null+&&+entrySet.size>0)+{
++++++++++++for(Entry<String,String>+entry+:+entrySet)+{
++++++++++++++++if(entry!=null+&&+paramName.equalsIgnoreCase(entry.getValue()))+{
++++++++++++++++++++keyForValue+=+entry.getKey();
++++++++++++++++}
++++++++++++}
++++++++}
++++}
++++return+keyForValue;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|33992|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>/**
 * This method gets the Key for the given Value
 * @param paramName
 * @return
 */
private String getKeyForValueFromMap(String paramName) {
 String keyForValue = null;
 if(paramName!=null)) {
 Set&lt;Entry&lt;String,String&gt;&gt; entrySet = myMap().entrySet();
 if(entrySet!=null &amp;&amp; entrySet.size&gt;0) {
 for(Entry&lt;String,String&gt; entry : entrySet) {
 if(entry!=null &amp;&amp; paramName.equalsIgnoreCase(entry.getValue())) {
 keyForValue = entry.getKey();
 }
 }
 }
 }
 return keyForValue;
}
</code></pre>

blocks|key|34038|text|需要注意的是，由于这个问题，Apache+Collections支持Generic+BidiMaps。因此，一些投票最多的答案在这一点上不再准确。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|34039|对于也支持重复值的序列化BidiMap+(一对多场景)，也可以考虑MapDB.org。|34040|entityMap|0|LINK|mutability|MUTABLE|url|https://commons.apache.org/proper/commons-collections/javadocs/api-release/org/apache/commons/collections4/bidimap/package-summary.html|1|https://github.com/jankotek/mapdb/blob/master/src/test/java/examples/Bidi_Map.java^0|Y|G|0|0|X|9|1|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]]|C|$]]|$1|D|3|E|5|6|7|T|8|@]|9|@$A|U|B|V|1|W]]|C|$]]|$1|F|3|-4|5|6|7|X|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]|N|$5|I|J|K|C|$L|O]]]]

It's important to note that since this question, Apache Collections supports <a href="https://commons.apache.org/proper/commons-collections/javadocs/api-release/org/apache/commons/collections4/bidimap/package-summary.html" rel="nofollow">Generic BidiMaps</a>. So a few of the top voted answers are no longer accurate on that point.

For a Serialized BidiMap that also supports duplicate values ( 1-to-many scenario ) also consider <a href="https://github.com/jankotek/mapdb/blob/master/src/test/java/examples/Bidi_Map.java" rel="nofollow">MapDB.org</a>.

blocks|key|34056|text|for(int+key:+hm.keySet())+{
++++if(hm.get(key).equals(value))+{
++++++++System.out.println(key);+
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|34057|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>for(int key: hm.keySet()) {
 if(hm.get(key).equals(value)) {
 System.out.println(key); 
 }
}
</code></pre>

blocks|key|34070|text|使用Java+8：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|34071|ftw.forEach((key,+value)+->+{
++++if+(value.equals("foo"))+{
++++++++System.out.print(key);
++++}
});|code-block|syntax|javascript|34072|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Using Java 8:

<pre><code>ftw.forEach((key, value) -&gt; {
 if (value.equals("foo")) {
 System.out.print(key);
 }
});
</code></pre>

blocks|key|282675|text|对于针对API+<+19的Android开发，Vitalii+Fedorenko一对一关系解决方案不起作用，因为Objects.equals没有实现。这里有一个简单的替代方案：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|282676|public+<K,+V>+K+getKeyByValue(Map<K,+V>+map,+V+value)+{
++++for+(Map.Entry<K,+V>+entry+:+map.entrySet())+{
++++++++++++if+(value.equals(entry.getValue()))+{
++++++++++++return+entry.getKey();
++++++++}
++++}
++++return+null;
}|code-block|syntax|javascript|282677|entityMap^0|1K|E|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

For Android development targeting API &lt; 19, Vitalii Fedorenko one-to-one relationship solution doesn't work because <code>Objects.equals</code> isn't implemented. Here's a simple alternative:

<pre><code>public &lt;K, V&gt; K getKeyByValue(Map&lt;K, V&gt; map, V value) {
 for (Map.Entry&lt;K, V&gt; entry : map.entrySet()) {
 if (value.equals(entry.getValue())) {
 return entry.getKey();
 }
 }
 return null;
}
</code></pre>

blocks|key|34115|text|使用您自己的实现来装饰map|type|unstyled|depth|inlineStyleRanges|entityRanges|data|34116|class+MyMap<K,V>+extends+HashMap<K,+V>{

++++Map<V,K>+reverseMap+=+new+HashMap<V,K>();

++++@Override
++++public+V+put(K+key,+V+value)+{
++++++++//+TODO+Auto-generated+method+stub
++++++++reverseMap.put(value,+key);
++++++++return+super.put(key,+value);
++++}

++++public+K+getKey(V+value){
++++++++return+reverseMap.get(value);
++++}
}|code-block|syntax|javascript|34117|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Decorate map with your own implementation

<pre><code>class MyMap&lt;K,V&gt; extends HashMap&lt;K, V&gt;{

 Map&lt;V,K&gt; reverseMap = new HashMap&lt;V,K&gt;();

 @Override
 public V put(K key, V value) {
 // TODO Auto-generated method stub
 reverseMap.put(value, key);
 return super.put(key, value);
 }

 public K getKey(V value){
 return reverseMap.get(value);
 }
}
</code></pre>

blocks|key|282698|text|我的两分钱。您可以获取数组中的键，然后循环遍历该数组。如果map非常大，这将影响此代码块的性能，在这种情况下，您首先获取数组中的键，这可能会消耗一些时间，然后再循环。否则，对于较小的地图，它应该是可以的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|282699|String[]+keys+=++yourMap.keySet().toArray(new+String[0]);

for(int+i+=+0+;+i+<+keys.length+;+i%2B%2B){
++++//This+is+your+key++++
++++String+key+=+keys[i];

++++//This+is+your+value
++++yourMap.get(key)++++++++++++
}|code-block|syntax|javascript|282700|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

My 2 cents. 
You can get the keys in an array and then loop through the array. This will affect performance of this code block if the map is pretty big , where in you are getting the keys in an array first which might consume some time and then you are looping. Otherwise for smaller maps it should be ok.

<pre><code>String[] keys = yourMap.keySet().toArray(new String[0]);

for(int i = 0 ; i &lt; keys.length ; i++){
 //This is your key 
 String key = keys[i];

 //This is your value
 yourMap.get(key) 
}
</code></pre>

blocks|key|1582060|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1582061|如果你想从value中获取密钥，最好使用双向映射(+bidimap+)，你可以在O(1)时间内从value中获取密钥。|ordered-list-item|1582062|1582063|但是，这样做的缺点是您只能使用唯一的键集，而valueset.|1582064|There是一种在java中称为表的数据结构，它只是如下所示的映射的映射|unordered-list-item|offset|length|style|BOLD|1582065|1582066|Table<+A，B，C>+==+map++>|1582067|在这里，您可以通过查询T.row(a);获得map<B,C>，也可以通过查询T.column(b);获得map<A,C>|CODE|1582068|在特殊情况下，将C作为某个常量插入。|1582069|因此，它就像<+a1，b1，1><+a2，b2，1>，...|1582070|所以，如果你通过T.row(a1)+->return+map+of+-->+get+keyset这个返回的map。|1582071|如果需要查找键值，则T.column(b2)+-->返回-->的映射-->获取返回映射的键集。|1582072|与前一种情况相比的优势：|1582073|1582074|可以使用多个值。|1582075|在使用大型数据集时效率更高。|1582076|1582077|entityMap^0|0|0|0|0|G|1|0|0|0|N|0|B|9|M|8|12|C|1G|8|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|1B|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|1C|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|1D|7|@]|8|@]|9|$]]|$1|E|3|F|4|5|6|1E|7|@]|8|@]|9|$]]|$1|G|3|H|4|I|6|1F|7|@$J|1G|K|1H|L|M]]|8|@]|9|$]]|$1|N|3|-4|4|5|6|1I|7|@]|8|@]|9|$]]|$1|O|3|P|4|5|6|1J|7|@$J|1K|K|1L|L|M]]|8|@]|9|$]]|$1|Q|3|R|4|5|6|1M|7|@$J|1N|K|1O|L|S]|$J|1P|K|1Q|L|S]|$J|1R|K|1S|L|S]|$J|1T|K|1U|L|S]]|8|@]|9|$]]|$1|T|3|U|4|5|6|1V|7|@]|8|@]|9|$]]|$1|V|3|W|4|5|6|1W|7|@]|8|@]|9|$]]|$1|X|3|Y|4|5|6|1X|7|@]|8|@]|9|$]]|$1|Z|3|10|4|5|6|1Y|7|@]|8|@]|9|$]]|$1|11|3|12|4|5|6|1Z|7|@]|8|@]|9|$]]|$1|13|3|-4|4|5|6|20|7|@]|8|@]|9|$]]|$1|14|3|15|4|C|6|21|7|@]|8|@]|9|$]]|$1|16|3|17|4|C|6|22|7|@]|8|@]|9|$]]|$1|18|3|-4|4|5|6|23|7|@]|8|@]|9|$]]|$1|19|3|-4|4|5|6|24|7|@]|8|@]|9|$]]]|1A|$]]

<ol>
<li>If you want to get key from value, its best to use bidimap (bi-directional maps) , you can get key from value in O(1) time.

But, the drawback with this is you can only use unique keyset and valueset.</li>
<li>There is a data structure called Table in java, which is nothing but map of maps like

Table&lt; A, B , C > == map &lt; A , map &lt; B, C > >

Here you can get <code>map&lt;B,C&gt;</code> by querying <code>T.row(a);</code>, and you can also get <code>map&lt;A,C&gt;</code> by querying <code>T.column(b);</code></li>
</ol>

In your special case, insert C as some constant.

So, it like &lt; a1, b1, 1 > 
 &lt; a2, b2 , 1 > , ...

So, if you find via T.row(a1) ---> returns map of --> get keyset this returned map.

If you need to find key value then, T.column(b2) --> returns map of --> get keyset of returned map.

Advantages over the previous case : 

<ol>
<li>Can use multiple values.</li>
<li>More efficient when using large data sets.</li>
</ol>

blocks|key|1438467|text|您可以使用以下代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1438468|public+class+HashmapKeyExist+{
++++public+static+void+main(String[]+args)+{
++++++++HashMap<String,+String>+hmap+=+new+HashMap<String,+String>();
++++++++hmap.put("1",+"Bala");
++++++++hmap.put("2",+"Test");

++++++++Boolean+cantain+=+hmap.containsValue("Bala");
++++++++if(hmap.containsKey("2")+&&+hmap.containsValue("Test"))
++++++++{
++++++++++++System.out.println("Yes");
++++++++}
++++++++if(cantain+==+true)
++++++++{
++++++++++++System.out.println("Yes");+
++++++++}

++++++++Set+setkeys+=+hmap.keySet();
++++++++Iterator+it+=+setkeys.iterator();

++++++++while(it.hasNext())
++++++++{
++++++++++++String+key+=+(String)+it.next();
++++++++++++if+(hmap.get(key).equals("Bala"))
++++++++++++{
++++++++++++++++System.out.println(key);
++++++++++++}
++++++++}
++++}
}|code-block|syntax|javascript|1438469|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can use the below:

<pre><code>public class HashmapKeyExist {
 public static void main(String[] args) {
 HashMap&lt;String, String&gt; hmap = new HashMap&lt;String, String&gt;();
 hmap.put("1", "Bala");
 hmap.put("2", "Test");

 Boolean cantain = hmap.containsValue("Bala");
 if(hmap.containsKey("2") &amp;&amp; hmap.containsValue("Test"))
 {
 System.out.println("Yes");
 }
 if(cantain == true)
 {
 System.out.println("Yes"); 
 }

 Set setkeys = hmap.keySet();
 Iterator it = setkeys.iterator();

 while(it.hasNext())
 {
 String key = (String) it.next();
 if (hmap.get(key).equals("Bala"))
 {
 System.out.println(key);
 }
 }
 }
}
</code></pre>

blocks|key|282734|text|public+static+String+getKey(Map<String,+Integer>+mapref,+String+value)+{
++++String+key+=+"";
++++for+(Map.Entry<String,+Integer>+map+:+mapref.entrySet())+{
++++++++if+(map.getValue().toString().equals(value))+{
++++++++++++key+=+map.getKey();
++++++++}
++++}
++++return+key;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|282735|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public static String getKey(Map&lt;String, Integer&gt; mapref, String value) {
 String key = "";
 for (Map.Entry&lt;String, Integer&gt; map : mapref.entrySet()) {
 if (map.getValue().toString().equals(value)) {
 key = map.getKey();
 }
 }
 return key;
}
</code></pre>

blocks|key|1584559|text|我认为keySet()可以很好地找到映射到值的键，并且拥有比entrySet()更好的编码风格。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|1584560|例如：|1584561|假设您有一个HashMap映射，ArrayList+res，一个要查找所有键映射的值，然后将键存储到res。|1584562|您可以编写以下代码：|1584563|++++for+(int+key+:+map.keySet())+{
++++++++if+(map.get(key)+==+value)+{
++++++++++++res.add(key);
++++++++}
++++}|code-block|syntax|javascript|1584564|而不是使用下面的entrySet()：|1584565|++++for+(Map.Entry+s+:+map.entrySet())+{
++++++++if+((int)s.getValue()+==+value)+{
++++++++++++res.add((int)s.getKey());
++++++++}
++++}|1584566|希望它能有所帮助:)|1584567|entityMap^0|3|8|U|A|0|0|15|1|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|13|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|17|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|18|8|@]|D|@]|E|$O|P]]|$1|Q|3|R|5|6|7|19|8|@]|D|@]|E|$]]|$1|S|3|T|5|N|7|1A|8|@]|D|@]|E|$O|P]]|$1|U|3|V|5|6|7|1B|8|@]|D|@]|E|$]]|$1|W|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|X|$]]

I think keySet() may be well to find the keys mapping to the value, and have a better coding style than entrySet().

Ex:

Suppose you have a HashMap map, ArrayList res, a value you want to find all the key mapping to , then store keys to the res.

You can write code below:

<pre><code> for (int key : map.keySet()) {
 if (map.get(key) == value) {
 res.add(key);
 }
 }
</code></pre>

rather than use entrySet() below:

<pre><code> for (Map.Entry s : map.entrySet()) {
 if ((int)s.getValue() == value) {
 res.add((int)s.getKey());
 }
 }
</code></pre>

Hope it helps :)

blocks|key|34277|text|虽然这不能直接回答这个问题，但它是相关的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|34278|这样你就不需要一直创建/迭代了。只需创建一次反向映射，即可获得所需内容。|34279|/**
+*+Both+key+and+value+types+must+define+equals()+and+hashCode()+for+this+to+work.
+*+This+takes+into+account+that+all+keys+are+unique+but+all+values+may+not+be.
+*
+*+@param+map
+*+@param+<K>
+*+@param+<V>
+*+@return
+*/
public+static+<K,+V>+Map<V,+List<K>>+reverseMap(Map<K,V>+map)+{
++++if(map+==+null)+return+null;

++++Map<V,+List<K>>+reverseMap+=+new+ArrayMap<>();

++++for(Map.Entry<K,V>+entry+:+map.entrySet())+{
++++++++appendValueToMapList(reverseMap,+entry.getValue(),+entry.getKey());
++++}

++++return+reverseMap;
}


/**
+*+Takes+into+account+that+the+list+may+already+have+values.
+*+
+*+@param+map
+*+@param+key
+*+@param+value
+*+@param+<K>
+*+@param+<V>
+*+@return
+*/
public+static+<K,+V>+Map<K,+List<V>>+appendValueToMapList(Map<K,+List<V>>+map,+K+key,+V+value)+{
++++if(map+==+null+%7C%7C+key+==+null+%7C%7C+value+==+null)+return+map;

++++List<V>+list+=+map.get(key);

++++if(list+==+null)+{
++++++++List<V>+newList+=+new+ArrayList<>();
++++++++newList.add(value);
++++++++map.put(key,+newList);
++++}
++++else+{
++++++++list.add(value);
++++}

++++return+map;
}|code-block|syntax|javascript|34280|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

While this does not directly answer the question, it is related.

This way you don't need to keep creating/iterating. Just create a reverse map once and get what you need.

<pre><code>/**
 * Both key and value types must define equals() and hashCode() for this to work.
 * This takes into account that all keys are unique but all values may not be.
 *
 * @param map
 * @param &lt;K&gt;
 * @param &lt;V&gt;
 * @return
 */
public static &lt;K, V&gt; Map&lt;V, List&lt;K&gt;&gt; reverseMap(Map&lt;K,V&gt; map) {
 if(map == null) return null;

 Map&lt;V, List&lt;K&gt;&gt; reverseMap = new ArrayMap&lt;&gt;();

 for(Map.Entry&lt;K,V&gt; entry : map.entrySet()) {
 appendValueToMapList(reverseMap, entry.getValue(), entry.getKey());
 }

 return reverseMap;
}


/**
 * Takes into account that the list may already have values.
 * 
 * @param map
 * @param key
 * @param value
 * @param &lt;K&gt;
 * @param &lt;V&gt;
 * @return
 */
public static &lt;K, V&gt; Map&lt;K, List&lt;V&gt;&gt; appendValueToMapList(Map&lt;K, List&lt;V&gt;&gt; map, K key, V value) {
 if(map == null || key == null || value == null) return map;

 List&lt;V&gt; list = map.get(key);

 if(list == null) {
 List&lt;V&gt; newList = new ArrayList&lt;&gt;();
 newList.add(value);
 map.put(key, newList);
 }
 else {
 list.add(value);
 }

 return map;
}
</code></pre>

blocks|key|1582115|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1582116|static+String+getKeyFromValue(LinkedHashMap<String,+String>+map,String+value)+{
++++for+(int+x=0;x<map.size();x%2B%2B){
++++++++if(+String.valueOf(+(new+ArrayList<String>(map.values())).get(x)+).equals(value))
++++++++++++return+String.valueOf((new+ArrayList<String>(map.keySet())).get(x));
++++}
++++return+null;
}|code-block|syntax|javascript|1582117|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

try this:

<pre><code>static String getKeyFromValue(LinkedHashMap&lt;String, String&gt; map,String value) {
 for (int x=0;x&lt;map.size();x++){
 if( String.valueOf( (new ArrayList&lt;String&gt;(map.values())).get(x) ).equals(value))
 return String.valueOf((new ArrayList&lt;String&gt;(map.keySet())).get(x));
 }
 return null;
}
</code></pre>

blocks|key|1582124|text|据我所知，当您将HashMap的键和值表示为数组时，它们不会混合在一起：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1582125|hashmap.values().toArray()|code-block|syntax|javascript|1582126|和|1582127|hashmap.keySet().toArray()|1582128|因此，下面的代码(从java+8开始)应该可以按预期工作：|1582129|public+Object+getKeyByFirstValue(Object+value)+{
++++int+keyNumber+=++Arrays.asList(hashmap.values().toArray()).indexOf(value);
++++return+hashmap.keySet().toArray()[keyNumber];
}|1582130|然而，(警告！)它的工作速度比迭代慢2-3倍。|offset|length|style|BOLD|1582131|entityMap^0|0|0|0|0|0|0|4|3|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@$Q|13|R|14|S|T]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

As far as I know keys and values of a HashMap are not mixed when you represent them as arrays:

<pre><code>hashmap.values().toArray()
</code></pre>

and

<pre><code>hashmap.keySet().toArray()
</code></pre>

So the following code (since java 8) should work as expected:

<pre><code>public Object getKeyByFirstValue(Object value) {
 int keyNumber = Arrays.asList(hashmap.values().toArray()).indexOf(value);
 return hashmap.keySet().toArray()[keyNumber];
}
</code></pre>

However, (WARNING!) it works 2-3 times slower than iteration.

blocks|key|34331|text|设定值为maxValue。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|34332|Set+keySet+=+map.keySet();

keySet.stream().filter(x->map.get(x)==maxValue).forEach(x->+System.out.println(x));|code-block|syntax|javascript|34333|entityMap^0|4|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Let the value be <code>maxValue</code>.

<pre><code>Set keySet = map.keySet();

keySet.stream().filter(x-&gt;map.get(x)==maxValue).forEach(x-&gt; System.out.println(x));
</code></pre>

blocks|key|282781|text|lambda，不使用外部库|type|unstyled|depth|inlineStyleRanges|entityRanges|data|282782|一个键可以处理多个值(与BidiMap有所不同)|282783|public+static+List<String>+getKeysByValue(Map<String,+String>+map,+String+value)+{
++List<String>+list+=+map.keySet().stream()
++++++.collect(groupingBy(k+->+map.get(k))).get(value);
++return+(list+==+null+?+Collections.emptyList()+:+list);
}|code-block|syntax|javascript|282784|获取包含key(s)映射value的List|offset|length|style|CODE|282785|对于1:1映射，返回的列表为empty或包含1个value|282786|entityMap^0|0|0|0|4|6|C|5|I|4|0|E|5|O|5|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@$K|W|L|X|M|N]|$K|Y|L|Z|M|N]|$K|10|L|11|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|12|8|@$K|13|L|14|M|N]|$K|15|L|16|M|N]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$]]

lambda w/o use of external libraries 
can deal with multiple values for one key (in difference to the BidiMap)
<pre><code>public static List&lt;String&gt; getKeysByValue(Map&lt;String, String&gt; map, String value) {
 List&lt;String&gt; list = map.keySet().stream()
 .collect(groupingBy(k -&gt; map.get(k))).get(value);
 return (list == null ? Collections.emptyList() : list);
}
</code></pre>
gets a <code>List</code> containing the <code>key(s)</code> mapping <code>value</code> 
for an 1:1 mapping the returned list is <code>empty</code> or contains 1 <code>value</code>

blocks|key|282804|text|让我们看一下我的例子|type|unstyled|depth|inlineStyleRanges|entityRanges|data|282805|Map<String,+String>+mapPeopleAndCountry+=+new+HashMap<>();
mapPeopleAndCountry.put("Matis",+"Lithuania");
mapPeopleAndCountry.put("Carlos",+"Honduras");
mapPeopleAndCountry.put("Teboho",+"Lesotho");
mapPeopleAndCountry.put("Marielos",+"Honduras");


List<String>+peopleInHonduras+=+mapPeopleAndCountry.keySet()
++++.stream()
++++.filter(r+->+mapPeopleAndCountry.get(r)
++++++++++++++++.equals("Honduras"))
++++.stream(Collectors.toList());

//+will+return+["Carlos",+"Marielos"]|code-block|syntax|javascript|282806|注意:未经测试，可能包含打字错误|282807|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

let see my example
<pre class="lang-java prettyprint-override"><code>Map&lt;String, String&gt; mapPeopleAndCountry = new HashMap&lt;&gt;();
mapPeopleAndCountry.put(&quot;Matis&quot;, &quot;Lithuania&quot;);
mapPeopleAndCountry.put(&quot;Carlos&quot;, &quot;Honduras&quot;);
mapPeopleAndCountry.put(&quot;Teboho&quot;, &quot;Lesotho&quot;);
mapPeopleAndCountry.put(&quot;Marielos&quot;, &quot;Honduras&quot;);


List&lt;String&gt; peopleInHonduras = mapPeopleAndCountry.keySet()
 .stream()
 .filter(r -&gt; mapPeopleAndCountry.get(r)
 .equals(&quot;Honduras&quot;))
 .stream(Collectors.toList());

// will return [&quot;Carlos&quot;, &quot;Marielos&quot;]

</code></pre>
Note: untested, may contains typo

blocks|key|1438679|text|从Map中获取给定值的键的最简单实用方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1438680|public+static+void+fetchValue(Map<String,+Integer>+map,+Integer+i)
{+++
Stream+stream+=+map.entrySet().stream().filter(val->+val.getValue().equals(i)).map(Map.Entry::getKey);
stream.forEach(System.out::println);++++
}|code-block|syntax|javascript|1438681|详细说明：|1438682|1438683|方法fetchValue接受映射，该映射将字符串作为关键字，将整数作为值。|ordered-list-item|1438684|1438685|然后我们使用|1438686|().stream()将结果转换为流。|1438687|1438688|1438689|接下来我们使用filter+(中间操作)，它给我们一个等于第二个参数的值。|1438690|1438691|最后，我们使用forEach(|1438692|)来打印最终结果。|1438693|1438694|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|15|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|16|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|17|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|18|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|19|8|@]|9|@]|A|$]]|$1|P|3|Q|5|L|7|1A|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|T|3|U|5|L|7|1D|8|@]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1F|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|L|7|1G|8|@]|9|@]|A|$]]|$1|10|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]|$1|11|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]]|12|$]]

Simplest utility method to fetch a key of a given value from a Map:
<pre><code>public static void fetchValue(Map&lt;String, Integer&gt; map, Integer i)
{ 
Stream stream = map.entrySet().stream().filter(val-&gt; val.getValue().equals(i)).map(Map.Entry::getKey);
stream.forEach(System.out::println); 
}
</code></pre>
detailed explaination:
<ol>
<li>Method fetchValue accepts the map, which has String as key and Integer as value.
</li>
<li>Then we use entryset().stream() to convert result into a stream.
</li>
<li>Next we use filter (intermediate operation) which gives us a value that is equal to the second argument.
</li>
<li>Finally, we use forEach(final operation) to print our end result.
</li>
</ol>

blocks|key|1438733|text|找到的答案太多了。有些真的很棒。但我特别在寻找一种方法，这样我就可以使用循环来获得值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1438734|最后，我做了以下工作:对于HashMap一对一关系：|1438735|++++Map<String,+String>+map+=+new+HashMap<String,+String>();
++++
++++map.put("abc",+"123");
++++map.put("xyz",+"456");
++++
++++for(Entry<String,+String>+entry+:+map.entrySet())+{
++++++++if(entry.getValue().equalsIgnoreCase("456"))+{
++++++++++++System.out.println(entry.getKey());
++++++++}
++++}|code-block|syntax|javascript|1438736|输出："xyz“|1438737|对于HashMap一对多关系：|1438738|++++Map<String,+ArrayList<String>>+service+=+new+HashMap<String,+ArrayList<String>>();
++++
++++service.put("abc",+new+ArrayList<String>());
++++service.get("abc").add("a");
++++service.get("abc").add("b");
++++service.get("abc").add("c");
++++
++++service.put("xyz",+new+ArrayList<String>());
++++service.get("xyz").add("x");
++++service.get("xyz").add("y");
++++service.get("xyz").add("z");
++++
++++for(Entry<String,+ArrayList<String>>+entry+:+service.entrySet())+{
++++++++ArrayList<String>+values+=+entry.getValue();
++++++++for(String+value+:+values)+{
++++++++++++if(value.equalsIgnoreCase("x"))+{
++++++++++++++++System.out.println(entry.getKey());
++++++++++++}
++++++++}
++++++++
++++}|1438739|输出:+xyz|1438740|-Thanks|1438741|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|Z|8|@]|9|@]|A|$G|H]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

Found too many answers. Some were really great. But I was particularly looking for a way, so that I can get the value using loops.
So here is finally what I did:
For a HashMap 1-to-1 relation:
<pre><code> Map&lt;String, String&gt; map = new HashMap&lt;String, String&gt;();
 
 map.put(&quot;abc&quot;, &quot;123&quot;);
 map.put(&quot;xyz&quot;, &quot;456&quot;);
 
 for(Entry&lt;String, String&gt; entry : map.entrySet()) {
 if(entry.getValue().equalsIgnoreCase(&quot;456&quot;)) {
 System.out.println(entry.getKey());
 }
 }
</code></pre>
Output: &quot;xyz&quot;
For a HashMap 1-to-many relation:
<pre><code> Map&lt;String, ArrayList&lt;String&gt;&gt; service = new HashMap&lt;String, ArrayList&lt;String&gt;&gt;();
 
 service.put(&quot;abc&quot;, new ArrayList&lt;String&gt;());
 service.get(&quot;abc&quot;).add(&quot;a&quot;);
 service.get(&quot;abc&quot;).add(&quot;b&quot;);
 service.get(&quot;abc&quot;).add(&quot;c&quot;);
 
 service.put(&quot;xyz&quot;, new ArrayList&lt;String&gt;());
 service.get(&quot;xyz&quot;).add(&quot;x&quot;);
 service.get(&quot;xyz&quot;).add(&quot;y&quot;);
 service.get(&quot;xyz&quot;).add(&quot;z&quot;);
 
 for(Entry&lt;String, ArrayList&lt;String&gt;&gt; entry : service.entrySet()) {
 ArrayList&lt;String&gt; values = entry.getValue();
 for(String value : values) {
 if(value.equalsIgnoreCase(&quot;x&quot;)) {
 System.out.println(entry.getKey());
 }
 }
 
 }
</code></pre>
Output: xyz
-Thanks

blocks|key|34481|text|您也可以这样做:+First+:+put+map+(+key，value)+Second+:为了更新键，您需要删除表达式Third+:并使用oldValue放入一个新键|type|unstyled|depth|inlineStyleRanges|entityRanges|data|34482|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You can also do that work:
First : put map (key, value)
Second : to update the key you need to remove expression
Third : and put a new key with the oldValue

If I have the value <code>"foo"</code>, and a <code>HashMap&lt;String&gt; ftw</code> for which <code>ftw.containsValue("foo")</code> returns <code>true</code>, how can I get the corresponding key? Do I have to loop through the hashmap? What is the best way to do that?

Java Hashmap: How to get key from value?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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如果我有一个值"foo"，以及一个ftw.containsValue("foo")返回true的HashMap<String> ftw，我怎么能得到对应的键呢？我必须遍历hashmap吗？做到这一点的最好方法是什么？

问Java Hashmap:如何从值中获取键？
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问Java Hashmap:如何从值中获取键？EN