blocks|key|1409506|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1409507|String[]+trimmedArray+=+new+String[array.length];
for+(int+i+=+0;+i+<+array.length;+i%2B%2B)
++++trimmedArray[i]+=+array[i].trim();|code-block|syntax|javascript|1409508|现在trimmedArray包含与array相同的字符串，但没有前导空格和尾随空格。或者，您可以编写以下代码来就地修改同一数组中的字符串：|offset|length|style|CODE|1409509|for+(int+i+=+0;+i+<+array.length;+i%2B%2B)
++++array[i]+=+array[i].trim();|1409510|entityMap^0|0|0|2|C|H|5|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]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|P|$]]

Try this:

<pre><code>String[] trimmedArray = new String[array.length];
for (int i = 0; i &lt; array.length; i++)
 trimmedArray[i] = array[i].trim();
</code></pre>

Now <code>trimmedArray</code> contains the same strings as <code>array</code>, but without leading and trailing whitespace. Alternatively, you could write this for modifying the strings in-place in the same array:

<pre><code>for (int i = 0; i &lt; array.length; i++)
 array[i] = array[i].trim();
</code></pre>

blocks|key|1409597|text|另一个java+8+lambda选项：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1409598|String[]+array2+=+Arrays.stream(array).map(String::trim).toArray(String[]::new);|code-block|syntax|javascript|1409599|和丑陋但优化的版本，没有创建新的数组|1409600|Arrays.stream(array).map(String::trim).toArray(unused+->+array);|1409601|修改了原来的array。|1409602|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Another java 8 lambda option : 

<pre><code>String[] array2 = Arrays.stream(array).map(String::trim).toArray(String[]::new);
</code></pre>

And the ugly but optimized version without new array creation

<pre><code>Arrays.stream(array).map(String::trim).toArray(unused -&gt; array);
</code></pre>

Original "array" is modified.

blocks|key|5450782|text|在应用程序构建路径中添加commons-lang3-3.1.jar。使用下面的代码片段来裁剪字符串数组。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5450783|String+array+=+{"+String",+"Tom+Selleck+","+Fish+"};
array+=+StringUtils.stripAll(array);|code-block|syntax|javascript|5450784|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|$]]

Add commons-lang3-3.1.jar in your application build path.
Use the below code snippet to trim the String array.

<pre><code>String array = {" String", "Tom Selleck "," Fish "};
array = StringUtils.stripAll(array);
</code></pre>

blocks|key|1409624|text|在Java8中，Arrays.parallelSetAll似乎已经为此目的做好了准备：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1409625|import+java.util.Arrays;

Arrays.parallelSetAll(array,+(i)+->+array[i].trim());|code-block|syntax|javascript|1409626|这将就地修改原始数组，将每个元素替换为lambda表达式的结果。|1409627|entityMap^0|8|L|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

In Java 8, <code>Arrays.parallelSetAll</code> seems ready made for this purpose:

<pre><code>import java.util.Arrays;

Arrays.parallelSetAll(array, (i) -&gt; array[i].trim());
</code></pre>

This will modify the original array in place, replacing each element with the result of the lambda expression.

blocks|key|1953478|text|我知道这是一个非常古老的帖子，但从Java+1.8开始，有一种更好的方法来修剪数组中的每个字符串。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1953479|Java+8+Lamda表达式解决方案：|1953480|List<String>+temp+=+new+ArrayList<>(Arrays.asList(yourArray));
temp.forEach(e+->+{temp.set((temp.indexOf(e),+e.trim()});
yourArray+=+temp.toArray(new+String[temp.size()]);|code-block|syntax|javascript|1953481|使用此解决方案，您不必创建新的Array。|1953482|就像在斯卡·洛佩斯的解决方案中一样|1953483|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I know this is a really old post, but since Java 1.8 there is a nicer way to trim every String in an array.

Java 8 Lamda Expression solution:

<pre><code>List&lt;String&gt; temp = new ArrayList&lt;&gt;(Arrays.asList(yourArray));
temp.forEach(e -&gt; {temp.set((temp.indexOf(e), e.trim()});
yourArray = temp.toArray(new String[temp.size()]);
</code></pre>

with this solution you don't have to create a new Array. 
 Like in Óscar López's solution

blocks|key|5450500|text|您可以遍历数组中的元素并对每个元素调用array[i].trim()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5450501|entityMap^0|J|F|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|$]]

You can just iterate over the elements in the array and call <code>array[i].trim()</code> on each element

blocks|key|1409756|text|对于那些(像我一样)在Kotlin中寻找相同解决方案并仅指向Java的人-如何在Kotlin中修剪：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1409757|fun+main(args:+Array<String>)+{
++++//+array+definition
++++val+array+=+arrayListOf<String>("+String",+"Tom+Selleck+","+Fish+")
++++println(array)+//+print+original+->+[+String,+Tom+Selleck+,++Fish+]

++++//+remove+leading+and+trailing+spaces,+result+is+arrayList
++++val+sol1+=+array.map+{+it.trim()+}
++++println("sol1+=+$sol1")+//+->+sol1+=+[String,+Tom+Selleck,+Fish]

++++//+remove+leading+and+trailing+spaces,+result+is+String
++++val+sol2+=+array.joinToString+{+it.trim()+}
++++println("sol2+=+$sol2")+//+->+sol2+=+String,+Tom+Selleck,+Fish
}|code-block|syntax|javascript|1409758|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|$]]

For those (like me) who was looking for the same solution in Kotlin and were pointed to Java only - how to trim in Kotlin:
<pre><code>fun main(args: Array&lt;String&gt;) {
 // array definition
 val array = arrayListOf&lt;String&gt;(&quot; String&quot;, &quot;Tom Selleck &quot;,&quot; Fish &quot;)
 println(array) // print original -&gt; [ String, Tom Selleck , Fish ]

 // remove leading and trailing spaces, result is arrayList
 val sol1 = array.map { it.trim() }
 println(&quot;sol1 = $sol1&quot;) // -&gt; sol1 = [String, Tom Selleck, Fish]

 // remove leading and trailing spaces, result is String
 val sol2 = array.joinToString { it.trim() }
 println(&quot;sol2 = $sol2&quot;) // -&gt; sol2 = String, Tom Selleck, Fish
}
</code></pre>

blocks|key|5450808|text|不知道一开始(6年前)+OP是如何将{"+String",+"Tom+Selleck+","+Fish+"}放在一个数组中的，我想我应该分享我最终得到的东西。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5450809|我的数组是在字符串上使用split的结果，该字符串的分隔符周围可能有额外的空格。我的解决方案是在分裂的时候解决这个问题。我的代码如下。在测试之后，我将splitWithTrim()放入我的项目的Utils类中。它处理我的用例；如果您决定使用它，您可能需要考虑您可能会遇到什么样的字符串和分隔符。|5450810|public+class+Test+{
++++public+static+void+main(String[]+args)+{
++++++++test("+abc+def++ghi+++jkl+",+"+");
++++++++test("++abc;+def++;ghi++;+jkl;+",+";");
++++}

++++public+static+void+test(String+str,+String+splitOn)+{
++++++++System.out.println("Splitting+\""+%2B+str+%2B+"\"+on+\""+%2B+splitOn+%2B+"\"");
++++++++String[]+parts+=+splitWithTrim(str,+splitOn);
++++++++for+(String+part+:+parts)+{
++++++++++++System.out.println("("+%2B+part+%2B+")");
++++++++}
++++}

++++public+static+String[]+splitWithTrim(String+str,+String+splitOn)+{
++++++++if+(splitOn.equals("+"))+{
++++++++++++return+str.trim().split("+%2B");
++++++++}+else+{
++++++++++++return+str.trim().split("+*"+%2B+splitOn+%2B+"+*");
++++++++}
++++}
}|code-block|syntax|javascript|5450811|运行测试应用程序的输出是：|5450812|Splitting+"+abc+def++ghi+++jkl+"+on+"+"
(abc)
(def)
(ghi)
(jkl)
Splitting+"++abc;+def++;ghi++;+jkl;+"+on+";"
(abc)
(def)
(ghi)
(jkl)|5450813|entityMap^0|I|10|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

Not knowing how the OP happened to have <code>{" String", "Tom Selleck "," Fish "}</code> in an array in the first place (6 years ago), I thought I'd share what I ended up with.

My array is the result of using split on a string which might have extra spaces around delimiters. My solution was to address this at the point of the split. My code follows. After testing, I put splitWithTrim() in my Utils class of my project. It handles my use case; you might want to consider what sorts of strings and delimiters you might encounter if you decide to use it.

<pre><code>public class Test {
 public static void main(String[] args) {
 test(" abc def ghi jkl ", " ");
 test(" abc; def ;ghi ; jkl; ", ";");
 }

 public static void test(String str, String splitOn) {
 System.out.println("Splitting \"" + str + "\" on \"" + splitOn + "\"");
 String[] parts = splitWithTrim(str, splitOn);
 for (String part : parts) {
 System.out.println("(" + part + ")");
 }
 }

 public static String[] splitWithTrim(String str, String splitOn) {
 if (splitOn.equals(" ")) {
 return str.trim().split(" +");
 } else {
 return str.trim().split(" *" + splitOn + " *");
 }
 }
}
</code></pre>

Output of running the test application is:

<pre><code>Splitting " abc def ghi jkl " on " "
(abc)
(def)
(ghi)
(jkl)
Splitting " abc; def ;ghi ; jkl; " on ";"
(abc)
(def)
(ghi)
(jkl)
</code></pre>

blocks|key|5450587|text|String+val+=+"hi+hello+prince";
String+arr[]+=+val.split("+");

for+(int+i+=+0;+i+<+arr.length;+i%2B%2B)
{+++
+++++System.out.print(arr[i]);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5450588|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>String val = "hi hello prince";
String arr[] = val.split(" ");

for (int i = 0; i &lt; arr.length; i++)
{ 
 System.out.print(arr[i]);
}
</code></pre>

I was just wondering what the best way to remove the white space from all the elements of a list would be.

For example if I had <code>String [] array = {" String", "Tom Selleck "," Fish "}</code>
How could I get all the elements as <code>{"String","Tom Selleck","Fish"}</code>

Thanks!

How to trim white space from all elements in array?

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。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我想知道从列表的所有元素中删除空格的最好方法是什么。例如，如果我有String [] array = {" String", "Tom Selleck "," Fish "}，如何才能获得{"String","Tom Selleck","Fish"}形式的所有元素谢谢!

问如何从数组中的所有元素中裁剪空白？
EN

回答 9

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何从数组中的所有元素中裁剪空白？EN

回答 9

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何从数组中的所有元素中裁剪空白？
EN