blocks|key|3099964|text|将数组保存在局部变量中，并使用数组的length字段查找其长度。以0为基数减去1：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3099965|String[]+bits+=+one.split("-");
String+lastOne+=+bits[bits.length-1];|code-block|syntax|javascript|3099966|注意:如果原始字符串仅由分隔符组成，例如"-"或"---"，则bits.length将为0，这将抛出ArrayIndexOutOfBoundsException。示例：https://onlinegdb.com/r1M-TJkZ8|3099967|entityMap|0|LINK|mutability|MUTABLE|url|https://onlinegdb.com/r1M-TJkZ8^0|I|6|0|0|K|3|O|5|V|B|2C|V|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]]|E|$]]|$1|M|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

Save the array in a local variable and use the array's <code>length</code> field to find its length. Subtract one to account for it being 0-based:

<pre><code>String[] bits = one.split("-");
String lastOne = bits[bits.length-1];
</code></pre>

Caveat emptor: if the original string is composed of only the separator, for example <code>"-"</code> or <code>"---"</code>, <code>bits.length</code> will be 0 and this will throw an ArrayIndexOutOfBoundsException. Example: <a href="https://onlinegdb.com/r1M-TJkZ8" rel="noreferrer">https://onlinegdb.com/r1M-TJkZ8</a>

blocks|key|335040|text|或者，您可以对字符串使用lastIndexOf()方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|335041|String+last+=+string.substring(string.lastIndexOf('-')+%2B+1);|code-block|syntax|javascript|335042|entityMap^0|C|D|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|$]]

Or you could use <code>lastIndexOf()</code> method on String

<pre><code>String last = string.substring(string.lastIndexOf('-') + 1);
</code></pre>

blocks|key|3100105|text|您可以在Apache+Commons中使用StringUtils类：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3100106|StringUtils.substringAfterLast(one,+"-");|code-block|syntax|javascript|3100107|entityMap|0|LINK|mutability|MUTABLE|url|https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#substringAfterLast-java.lang.String-java.lang.String-^0|L|B|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]]]]

You can use the <a href="https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#substringAfterLast-java.lang.String-java.lang.String-" rel="noreferrer">StringUtils</a> class in Apache Commons: 

<pre><code>StringUtils.substringAfterLast(one, "-");
</code></pre>

blocks|key|333605|text|使用一个简单但通用的助手方法，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|333606|public+static+<T>+T+last(T[]+array)+{
++++return+array[array.length+-+1];
}|code-block|syntax|javascript|333607|你可以重写：|333608|lastone+=+one.split("-")[..];|333609|作为：|333610|lastone+=+last(one.split("-"));|333611|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

using a simple, yet generic, helper method like this:

<pre><code>public static &lt;T&gt; T last(T[] array) {
 return array[array.length - 1];
}
</code></pre>

you can rewrite:

<pre><code>lastone = one.split("-")[..];
</code></pre>

as:

<pre><code>lastone = last(one.split("-"));
</code></pre>

blocks|key|333628|text|String+str+=+"www.anywebsite.com/folder/subfolder/directory";
int+index+=+str.lastIndexOf('/');
String+lastString+=+str.substring(index+%2B1);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|333629|现在，lastString的值为"directory"|unstyled|offset|length|style|CODE|333630|entityMap^0|0|3|A|G|B|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]|$G|Q|H|R|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|S|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>String str = "www.anywebsite.com/folder/subfolder/directory";
int index = str.lastIndexOf('/');
String lastString = str.substring(index +1);
</code></pre>

Now <code>lastString</code> has the value <code>"directory"</code>

blocks|key|3312947|text|使用Guava|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3312948|final+Splitter+splitter+=+Splitter.on("-").trimResults();
assertEquals("Günnewig+Uebachs",+Iterables.getLast(splitter.split(one)));
assertEquals("Madison",+Iterables.getLast(splitter.split(two)));|code-block|syntax|javascript|3312949|Splitter，Iterables|style|CODE|3312950|entityMap|0|LINK|mutability|MUTABLE|url|https://code.google.com/p/guava-libraries/|1|http://docs.guava-libraries.googlecode.com/git-history/release/javadoc/com/google/common/base/Splitter.html|2|http://docs.guava-libraries.googlecode.com/git/javadoc/com/google/common/collect/Iterables.html^0|2|5|0|0|0|0|8|9|9|0|8|1|9|9|2|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|13|8|@$A|14|B|15|K|L]|$A|16|B|17|K|L]]|9|@$A|18|B|19|1|1A]|$A|1B|B|1C|1|1D]]|C|$]]|$1|M|3|-4|5|6|7|1E|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]|W|$5|P|Q|R|C|$S|X]]]]

With <a href="https://code.google.com/p/guava-libraries/" rel="noreferrer">Guava</a>:

<pre><code>final Splitter splitter = Splitter.on("-").trimResults();
assertEquals("Günnewig Uebachs", Iterables.getLast(splitter.split(one)));
assertEquals("Madison", Iterables.getLast(splitter.split(two)));
</code></pre>

<a href="http://docs.guava-libraries.googlecode.com/git-history/release/javadoc/com/google/common/base/Splitter.html" rel="noreferrer"><code>Splitter</code></a>, <a href="http://docs.guava-libraries.googlecode.com/git/javadoc/com/google/common/collect/Iterables.html" rel="noreferrer"><code>Iterables</code></a>

blocks|key|3100067|text|由于他要求使用split在同一行中完成所有这些操作，因此我建议这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3100068|lastone+=+one.split("-")[(one.split("-")).length+-1]++|code-block|syntax|javascript|3100069|我总是尽可能地避免定义新的变量，我发现这是一个非常好的实践|3100070|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|$]]

Since he was asking to do it all in the same line using split so i suggest this:

<pre><code>lastone = one.split("-")[(one.split("-")).length -1] 
</code></pre>

I always avoid defining new variables as far as I can, and I find it a very good practice

blocks|key|338505|text|你的意思是你不知道编译时数组的大小？在运行时，可以通过lastone.length和lastwo.length的值找到它们。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|338506|entityMap^0|R|E|16|D|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|$]]

You mean you don't know the sizes of the arrays at compile-time? At run-time they could be found by the value of <code>lastone.length</code> and <code>lastwo.length</code> .

blocks|key|336241|text|在java+8中|type|unstyled|depth|inlineStyleRanges|entityRanges|data|336242|String+lastItem+=+Stream.of(str.split("-")).reduce((first,last)->last).get();|code-block|syntax|javascript|336243|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 java 8 

<pre><code>String lastItem = Stream.of(str.split("-")).reduce((first,last)-&gt;last).get();
</code></pre>

blocks|key|3313051|text|您也可以使用java.util.ArrayDeque|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3313052|String+last+=+new+ArrayDeque<>(Arrays.asList("1-2".split("-"))).getLast();|code-block|syntax|javascript|3313053|entityMap|0|LINK|mutability|MUTABLE|url|https://www.tutorialspoint.com/java/util/java_util_arraydeque.htm^0|6|K|6|K|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Also you can use <a href="https://www.tutorialspoint.com/java/util/java_util_arraydeque.htm" rel="nofollow noreferrer"><code>java.util.ArrayDeque</code></a>

<pre><code>String last = new ArrayDeque&lt;&gt;(Arrays.asList("1-2".split("-"))).getLast();
</code></pre>

blocks|key|3100151|text|我猜你想在I行中这样做。这是可能的(虽然有点杂乱无章=%5E)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3100152|new+StringBuilder(new+StringBuilder("Düsseldorf+-+Zentrum+-+Günnewig+Uebachs").reverse().toString().split("+-+")[0]).reverse()|code-block|syntax|javascript|3100153|tadaa，一行->你想要的结果(如果你拆分"-“(空格减去空格)而不只是”-“(减号)，你也会丢失分区前令人讨厌的空格=%5E)，所以"Günnewig+Uebach”而不是“Günnewig+Uebach”(以空格作为第一个字符)|3100154|漂亮的额外->库文件夹中不需要额外的JAR文件，因此您可以保持应用程序的轻量级。|3100155|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I guess you want to do this in i line. It is possible (a bit of juggling though =^)

<pre><code>new StringBuilder(new StringBuilder("Düsseldorf - Zentrum - Günnewig Uebachs").reverse().toString().split(" - ")[0]).reverse()
</code></pre>

tadaa, one line -> the result you want (if you split on " - " (space minus space) instead of only "-" (minus) you will loose the annoying space before the partition too =^) so "Günnewig Uebachs" instead of " Günnewig Uebachs" (with a space as first character)

Nice extra -> no need for extra JAR files in the lib folder so you can keep your application light weight.

I am using the String split method and I want to have the last element.
The size of the Array can change.

Example:

<pre><code>String one = "Düsseldorf - Zentrum - Günnewig Uebachs"
String two = "Düsseldorf - Madison"
</code></pre>

I want to split the above Strings and get the last item:

<pre><code>lastone = one.split("-")[here the last item] // &lt;- how?
lasttwo = two.split("-")[here the last item] // &lt;- how?
</code></pre>

I don't know the sizes of the arrays at runtime :(

Java: Get last element after split

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我正在使用字符串拆分方法，我想要最后一个元素。数组的大小可以更改。示例：String one = "Düsseldorf - Zentrum - Günnewig Uebachs"String two = "Düsseldorf - Madison"我想拆分上面的字符串并得到最后一项：lastone = one.split("-")[here the last item] // <- how?la

问Java:拆分后获取最后一个元素
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