blocks|key|1391442|text|您可以使用commons+lang的ArrayUtils。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1391443|array+=+ArrayUtils.removeElement(array,+element)|code-block|syntax|javascript|1391444|commons.apache.org+library:Javadocs|offset|length|1391445|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/ArrayUtils.html^0|0|0|0|Z|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

You could use commons lang's ArrayUtils.

<pre><code>array = ArrayUtils.removeElement(array, element)
</code></pre>

<a href="http://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/ArrayUtils.html" rel="noreferrer">commons.apache.org library:Javadocs</a>

blocks|key|1391537|text|你的问题不是很清楚。从你自己的回答中，我可以更好地告诉你你想要做什么：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1391538|public+static+String[]+removeElements(String[]+input,+String+deleteMe)+{
++++List+result+=+new+LinkedList();

++++for(String+item+:+input)
++++++++if(!deleteMe.equals(item))
++++++++++++result.add(item);

++++return+result.toArray(input);
}|code-block|syntax|javascript|1391539|注:这是未经测试的。错误检查留给读者作为练习(如果input或deleteMe为null，我将抛出IllegalArgumentException；null列表输入上的空列表没有任何意义。从数组中删除空字符串可能是有意义的，但我也将其作为练习；目前，如果deleteMe为空，则在尝试调用deleteMe上的equals时，它将抛出NPE。)|offset|length|style|CODE|1391540|我在这里做的选择：|1391541|我用的是LinkedList。迭代应该同样快，并且避免任何大小调整，或者如果最终删除了大量元素，则避免分配过大的列表。您可以使用ArrayList，并将初始大小设置为输入的长度。这可能不会有太大的不同。|1391542|entityMap^0|0|0|P|5|V|8|1D|O|3J|8|40|8|4M|3|0|0|4|A|1S|9|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]|$I|13|J|14|K|L]|$I|15|J|16|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|17|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|18|8|@$I|19|J|1A|K|L]|$I|1B|J|1C|K|L]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|R|$]]

Your question isn't very clear. From your own answer, I can tell better what you are trying to do:
<pre><code>public static String[] removeElements(String[] input, String deleteMe) {
 List result = new LinkedList();

 for(String item : input)
 if(!deleteMe.equals(item))
 result.add(item);

 return result.toArray(input);
}
</code></pre>
NB: This is untested. Error checking is left as an exercise to the reader (I'd throw <code>IllegalArgumentException</code> if either <code>input</code> or <code>deleteMe</code> is null; an empty list on null list input doesn't make sense. Removing null Strings from the array might make sense, but I'll leave that as an exercise too; currently, it will throw an <code>NPE</code> when it tries to call equals on <code>deleteMe</code> if <code>deleteMe</code> is null.)
Choices I made here:
I used a <code>LinkedList</code>. Iteration should be just as fast, and you avoid any resizes, or allocating too big of a list if you end up deleting lots of elements. You could use an <code>ArrayList</code>, and set the initial size to the length of input. It likely wouldn't make much of a difference.

blocks|key|1391082|text|不能从基本Java数组中删除元素。请看一下各种集合和ArrayList。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1391083|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't remove an element from the basic Java array. Take a look at various Collections and ArrayList instead.

blocks|key|1606515|text|你可以使用ArrayUtils+API以一种“漂亮的方式”移除它。它在数组上实现了许多操作(删除、查找、添加、包含等)。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1606516|瞧一瞧。它让我的生活变得更简单。|1606517|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/lang/api-release/org/apache/commons/lang3/ArrayUtils.html^0|5|E|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]]|C|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@]|C|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]]]

You could use the <a href="http://commons.apache.org/lang/api-release/org/apache/commons/lang3/ArrayUtils.html" rel="nofollow noreferrer">ArrayUtils API</a> to remove it in a "nice looking way". It implements many operations (remove, find, add, contains,etc) on Arrays. 
Take a look. It has made my life simpler.

blocks|key|1391410|text|好的，现在我经常这样用东西：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1391411|public+static+String[]+removeElements(String[]+input,+String+deleteMe)+{
++++if+(input+!=+null)+{
++++++++List<String>+list+=+new+ArrayList<String>(Arrays.asList(input));
++++++++for+(int+i+=+0;+i+<+list.size();+i%2B%2B)+{
++++++++++++if+(list.get(i).equals(deleteMe))+{
++++++++++++++++list.remove(i);
++++++++++++}
++++++++}
++++++++return+list.toArray(new+String[0]);
++++}+else+{
++++++++return+new+String[0];
++++}
}|code-block|syntax|javascript|1391412|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|$]]

okay, thx a lot
now i use sth like this:

<pre><code>public static String[] removeElements(String[] input, String deleteMe) {
 if (input != null) {
 List&lt;String&gt; list = new ArrayList&lt;String&gt;(Arrays.asList(input));
 for (int i = 0; i &lt; list.size(); i++) {
 if (list.get(i).equals(deleteMe)) {
 list.remove(i);
 }
 }
 return list.toArray(new String[0]);
 } else {
 return new String[0];
 }
}
</code></pre>

blocks|key|1391688|text|对于Bill+K和dadinn写的那些，需要一些更多的前提条件|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1391689|Object[]+newArray+=+new+Object[src.length+-+1];
if+(i+>+0){
++++System.arraycopy(src,+0,+newArray,+0,+i);
}

if+(newArray.length+>+i){
++++System.arraycopy(src,+i+%2B+1,+newArray,+i,+newArray.length+-+i);
}

return+newArray;|code-block|syntax|javascript|1391690|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|$]]

Some more pre-conditions are needed for the ones written by Bill K and dadinn

<pre><code>Object[] newArray = new Object[src.length - 1];
if (i &gt; 0){
 System.arraycopy(src, 0, newArray, 0, i);
}

if (newArray.length &gt; i){
 System.arraycopy(src, i + 1, newArray, i, newArray.length - i);
}

return newArray;
</code></pre>

blocks|key|1391759|text|您不能更改数组的长度，但可以通过复制新值并将其存储到现有索引号来更改索引包含的值。1=mike，2=jeff+//+10+=+george+11去1覆盖麦克.|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1391760|Object[]+array+=+new+Object[10];
int+count+=+-1;

public+void+myFunction(String+string)+{
++++count%2B%2B;
++++if(count+==+array.length)+{+
++++++++count+=+0;++//+overwrite+first
++++}
++++array[count]+=+string;++++
}|code-block|syntax|javascript|1391761|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 not change the length of an array, but you can change the values the index holds by copying new values and store them to a existing index number.
1=mike , 2=jeff // 10 = george 11 goes to 1 overwriting mike .

<pre><code>Object[] array = new Object[10];
int count = -1;

public void myFunction(String string) {
 count++;
 if(count == array.length) { 
 count = 0; // overwrite first
 }
 array[count] = string; 
}
</code></pre>

blocks|key|1606035|text|我希望您使用java集合/+java+commons集合！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1606036|使用java.util.ArrayList，您可以执行以下操作：|1606037|yourArrayList.remove(someObject);

yourArrayList.add(someObject);|code-block|syntax|javascript|1606038|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|$]]

I hope you use the java collection / java commons collections!

With an java.util.ArrayList you can do things like the following: 

<pre><code>yourArrayList.remove(someObject);

yourArrayList.add(someObject);
</code></pre>

blocks|key|1606103|text|将原始数组Copy到另一个数组中，而不删除要删除的元素。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1606104|一种更简单的方法是使用列表、集合...并使用remove()方法。|1606105|entityMap|0|LINK|mutability|MUTABLE|url|http://java.sun.com/j2se/1.5.0/docs/api/java/lang/System.html#arraycopy(java.lang.Object,%2520int,%2520java.lang.Object,%2520int,%2520int)^0|5|4|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]]|C|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@]|C|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]]]

<a href="http://java.sun.com/j2se/1.5.0/docs/api/java/lang/System.html#arraycopy(java.lang.Object,%20int,%20java.lang.Object,%20int,%20int)" rel="nofollow noreferrer">Copy</a> your original array into another array, without the element to be removed.

A simplier way to do that is to use a List, Set... and use the remove() method.

blocks|key|1391338|text|如果向下调整数组大小不感兴趣，请将要删除的项与最后一项互换。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1391339|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|$]]

Swap the item to be removed with the last item, if resizing the array down is not an interest.

blocks|key|1605978|text|使用ArrayList|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1605979|alist.remove(1);+//removes+the+element+at+position+1|code-block|syntax|javascript|1605980|entityMap^0|2|9|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|$]]

Use an <code>ArrayList</code>:

<pre><code>alist.remove(1); //removes the element at position 1
</code></pre>

blocks|key|1391103|text|当然，创建另一个数组:)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1391104|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|$]]

Is there any fast (and nice looking) way to remove an element from an array in Java?

Removing an element from an Array (Java)

Java

在Java中，有没有什么快速(且美观)的方法可以从数组中删除元素？

问从数组中删除元素(Java)
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回答 12

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