blocks|key|562752|text|使用String.trim()方法去掉空格(空格、换行符等)从字符串的开头和结尾开始。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|562753|String+trimmedString+=+myString.trim();|code-block|syntax|javascript|562754|entityMap|0|LINK|mutability|MUTABLE|url|http://download.oracle.com/javase/6/docs/api/java/lang/String.html#trim()^0|2|D|2|D|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]]]]

Use <a href="http://download.oracle.com/javase/6/docs/api/java/lang/String.html#trim()" rel="noreferrer"><code>String.trim()</code></a> method to get rid of whitespaces (spaces, new lines etc.) from the beginning and end of the string.

<pre><code>String trimmedString = myString.trim();
</code></pre>

blocks|key|562756|text|String.replaceAll("[\n\r]",+"");|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|562757|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.replaceAll("[\n\r]", "");
</code></pre>

blocks|key|562844|text|此Java代码执行的操作与问题标题中的要求完全相同，即“从字符串的开头和结尾删除换行符-java”：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|562845|String.replaceAll("%5E[\n\r]",+"").replaceAll("[\n\r]$",+"")|code-block|syntax|javascript|562846|仅从行尾删除换行符：|562847|String.replaceAll("[\n\r]$",+"")|562848|仅从行首删除换行符：|562849|String.replaceAll("%5E[\n\r]",+"")|562850|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|$]]

This Java code does exactly what is asked in the title of the question, that is "remove newlines from beginning and end of a string-java":

<pre><code>String.replaceAll("^[\n\r]", "").replaceAll("[\n\r]$", "")
</code></pre>

Remove newlines only from the end of the line:

<pre><code>String.replaceAll("[\n\r]$", "")
</code></pre>

Remove newlines only from the beginning of the line:

<pre><code>String.replaceAll("^[\n\r]", "")
</code></pre>

blocks|key|779290|text|如果您的字符串可能为null，请考虑使用StringUtils.trim()+-+String.trim()的空安全版本。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|779291|entityMap|0|LINK|mutability|MUTABLE|url|https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#trim-java.lang.String-|1|https://docs.oracle.com/javase/8/docs/api/java/lang/String.html#trim())^0|A|4|K|I|15|D|K|I|0|15|D|1|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@$9|W|A|X|1|Y]|$9|Z|A|10|1|11]]|E|$]]|$1|F|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]|N|$5|I|J|K|E|$L|O]]]]

If your string is potentially <code>null</code>, consider using <a href="https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#trim-java.lang.String-" rel="noreferrer"><code>StringUtils.trim()</code></a> - the null-safe version of <a href="https://docs.oracle.com/javase/8/docs/api/java/lang/String.html#trim()" rel="noreferrer"><code>String.trim()</code></a>.

blocks|key|562864|text|尝尝这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|562865|function+replaceNewLine(str)+{+
++return+str.replace(/[\n\r]/g,+"");
}|code-block|syntax|javascript|562866|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>function replaceNewLine(str) { 
 return str.replace(/[\n\r]/g, &quot;&quot;);
}
</code></pre>

blocks|key|562882|text|另一个优雅的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|562883|String+myString+=+"\nLogbasex\n";
myString+=+org.apache.commons.lang3.StringUtils.strip(myString,+"\n");|code-block|syntax|javascript|562884|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|$]]

Another elegant solution.
<pre><code>String myString = &quot;\nLogbasex\n&quot;;
myString = org.apache.commons.lang3.StringUtils.strip(myString, &quot;\n&quot;);
</code></pre>

blocks|key|562931|text|对于在处理不同的换行符时寻找问题答案的其他人：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|562932|string.replaceAll("(\n%7C\r%7C\r\n)$",+"");+//+Java+7
string.replaceAll("\\R$",+"");++++++++++//+Java+8|code-block|syntax|javascript|562933|这应该完全删除最后一个换行符，并保留字符串中的所有其他空格，并适用于Unix+(\n)、Windows+(\r\n)和旧的Mac+(\r)换行符：https://stackoverflow.com/a/20056634、https://stackoverflow.com/a/49791415。"\\R"是在Java8的Pattern类中引入的匹配器：https://docs.oracle.com/javase/8/docs/api/java/util/regex/Pattern.html|offset|length|style|CODE|562934|这将通过以下测试：|562935|//+Windows:
value+=+"\r\n+test+\r\n+value+\r\n";
assertEquals("\r\n+test+\r\n+value+",+value.replaceAll("\\R$",+""));

//+Unix:
value+=+"\n+test+\n+value+\n";
assertEquals("\n+test+\n+value+",+value.replaceAll("\\R$",+""));

//+Old+Mac:
value+=+"\r+test+\r+value+\r";
assertEquals("\r+test+\r+value+",+value.replaceAll("\\R$",+""));|562936|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/20056634|1|https://stackoverflow.com/a/49791415|2|https://docs.oracle.com/javase/8/docs/api/java/util/regex/Pattern.html^0|0|0|43|5|21|10|0|32|10|1|4W|1Y|2|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]]|$1|G|3|H|5|6|7|14|8|@$I|15|J|16|K|L]]|9|@$I|17|J|18|1|19]|$I|1A|J|1B|1|1C]|$I|1D|J|1E|1|1F]]|A|$]]|$1|M|3|N|5|6|7|1G|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|1H|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]|Y|$5|T|U|V|A|$W|Z]]|10|$5|T|U|V|A|$W|11]]]]

For anyone else looking for answer to the question when dealing with different linebreaks:
<pre><code>string.replaceAll(&quot;(\n|\r|\r\n)$&quot;, &quot;&quot;); // Java 7
string.replaceAll(&quot;\\R$&quot;, &quot;&quot;); // Java 8
</code></pre>
This should remove exactly the last line break and preserve all other whitespace from string and work with Unix (\n), Windows (\r\n) and old Mac (\r) line breaks: <a href="https://stackoverflow.com/a/20056634">https://stackoverflow.com/a/20056634</a>, <a href="https://stackoverflow.com/a/49791415">https://stackoverflow.com/a/49791415</a>. <code>&quot;\\R&quot;</code> is matcher introduced in Java 8 in Pattern class: <a href="https://docs.oracle.com/javase/8/docs/api/java/util/regex/Pattern.html" rel="nofollow noreferrer">https://docs.oracle.com/javase/8/docs/api/java/util/regex/Pattern.html</a>
This passes these tests:
<pre><code>// Windows:
value = &quot;\r\n test \r\n value \r\n&quot;;
assertEquals(&quot;\r\n test \r\n value &quot;, value.replaceAll(&quot;\\R$&quot;, &quot;&quot;));

// Unix:
value = &quot;\n test \n value \n&quot;;
assertEquals(&quot;\n test \n value &quot;, value.replaceAll(&quot;\\R$&quot;, &quot;&quot;));

// Old Mac:
value = &quot;\r test \r value \r&quot;;
assertEquals(&quot;\r test \r value &quot;, value.replaceAll(&quot;\\R$&quot;, &quot;&quot;));
</code></pre>

blocks|key|779262|text|String+text+=+readFileAsString("textfile.txt");
text+=+text.replace("\n",+"").replace("\r",+"");|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|779263|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 text = readFileAsString("textfile.txt");
text = text.replace("\n", "").replace("\r", "");
</code></pre>

I have a string that contains some text followed by a blank line. What's the best way to keep the part with text, but remove the whitespace newline from the end?

How to remove newlines from beginning and end of a string?

Windows

Java

Unix

我有一个字符串，它包含一些文本，后跟一个空行。最好的方法是保留部分文本，但去掉末尾的空格换行符？

问如何删除字符串开头和结尾的换行符？
EN

回答 8

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问如何删除字符串开头和结尾的换行符？EN

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问如何删除字符串开头和结尾的换行符？
EN