blocks|key|4835963|text|这将删除s末尾的任何回车符和换行符的组合|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4835964|s+=+s.TrimEnd(new+char[]+{+'\r',+'\n'+});|code-block|syntax|javascript|4835965|编辑：或者正如JP友好地指出的那样，你可以更简洁地拼写为：|BOLD|4835966|s+=+s.TrimEnd('\r',+'\n');|4835967|entityMap^0|4|1|0|0|0|2|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|U|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|V|8|@$9|W|A|X|B|M]]|D|@]|E|$]]|$1|N|3|O|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|P|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|Q|$]]

This will trim off any combination of carriage returns and newlines from the end of <code>s</code>:

<pre><code>s = s.TrimEnd(new char[] { '\r', '\n' });
</code></pre>

Edit: Or as JP kindly points out, you can spell that more succinctly as:

<pre><code>s = s.TrimEnd('\r', '\n');
</code></pre>

blocks|key|3990478|text|这应该行得通。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3990479|var+tst+=+"12345\n\n\r\n\r\r";
var+res+=+tst.TrimEnd(+'\r',+'\n'+);|code-block|syntax|javascript|3990480|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|$]]

This should work ...

<pre><code>var tst = "12345\n\n\r\n\r\r";
var res = tst.TrimEnd( '\r', '\n' );
</code></pre>

blocks|key|3990550|text|String+temp+=+s.Replace("\r\n","").Trim();|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3990551|s是原始字符串。(注大写)|unstyled|offset|length|style|CODE|3990552|entityMap^0|0|0|1|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]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>String temp = s.Replace(&quot;\r\n&quot;,&quot;&quot;).Trim();
</code></pre>
<code>s</code> being the original string. (Note capitals)

blocks|key|1252161|text|如果您使用多个平台，则使用此方法会更安全。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1252162|value.TrimEnd(System.Environment.NewLine.ToCharArray());|code-block|syntax|javascript|1252163|它将考虑不同的换行符和回车符。|1252164|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 are using multiple platforms you are safer using this method.

<pre><code>value.TrimEnd(System.Environment.NewLine.ToCharArray());
</code></pre>

It will account for different newline and carriage-return characters.

blocks|key|1252179|text|s.TrimEnd();|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1252180|以上就是我从字符串末尾删除'\r\n‘所需的全部内容。|unstyled|1252181|在我看来，被看好的答案似乎是错误的。首先，当我尝试时它不起作用，其次，如果它起作用了，我希望s.TrimEnd(+'\r‘，'\n')只删除'\r’或'\n'，所以我必须在我的字符串上运行两次-一次是当'\n‘在结尾时，第二次是当’\r‘在结尾时(现在'\n’已经被删除了)。|1252182|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>s.TrimEnd();
</code></pre>

The above is all I needed to remove '\r\n' from the end of my string.

The upvoted answer seems wrong to me. Firstly, it didn't work when I tried, secondly, if it did work I would expect that s.TrimEnd('\r', '\n') would only remove either a '\r' or a '\n', so I'd have to run it over my string twice - once for when '\n' was at the end and the second time for when '\r' was at the end (now that the '\n' was removed).

blocks|key|4836000|text|如果总是有一个CRLF，那么：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4836001|myString+=+myString.Substring(0,+myString.Length+-+2);|code-block|syntax|javascript|4836002|如果它可能有，也可能没有，那么：|4836003|Regex+re+=+new+Regex("\r\n$");
re.Replace(myString,+"");|4836004|这两个(根据设计)最多只能删除一个CRLF。缓存正则表达式以提高性能。|4836005|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|$]]

If there's always a single CRLF, then:

<pre><code>myString = myString.Substring(0, myString.Length - 2);
</code></pre>

If it may or may not have it, then:

<pre><code>Regex re = new Regex("\r\n$");
re.Replace(myString, "");
</code></pre>

Both of these (by design), will remove at most a single CRLF. Cache the regex for performance.

blocks|key|772045|text|对于我们的VBers：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|772046|TrimEnd(New+Char()+{ControlChars.Cr,+ControlChars.Lf})|code-block|syntax|javascript|772047|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 us VBers:

<pre><code>TrimEnd(New Char() {ControlChars.Cr, ControlChars.Lf})
</code></pre>

blocks|key|772053|text|string+k+=+"This+is+my\r\nugly+string.+I+want\r\nto+change+this.+Please+\r\n+help!";
k+=+System.Text.RegularExpressions.Regex.Replace(k,+@"\r\n%2B",+"+");|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|772054|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 k = "This is my\r\nugly string. I want\r\nto change this. Please \r\n help!";
k = System.Text.RegularExpressions.Regex.Replace(k, @"\r\n+", " ");
</code></pre>

blocks|key|772085|text|这太简单了--对我来说，我正在过滤掉某些电子邮件项目。我正在编写自己的自定义电子邮件垃圾过滤器。如果字符串中有\r和/或\n，它将清除所有项目，而不是过滤。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|772086|所以，我只做了filter+=+filter.Remove('\n')和filter+=+filter.Remove('\r')。我正在制作过滤器，以便最终用户可以使用记事本直接编辑文件，因此不知道这些字符可能嵌入到哪里--可能不在字符串的开头或结尾。所以把它们都去掉就行了。|772087|其他条目都可以工作，但删除可能是最简单的？|772088|我从这篇文章中学到了更多关于Regex的知识--在这里使用它非常酷。|772089|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

This was too easy -- for me I'm filtering out certain email items. I'm writing my own custom email junk filter. With \r and/or \n in the string it was wiping out all items instead of filtering.

So, I just did filter = filter.Remove('\n') and filter = filter.Remove('\r'). I'm making my filter such that an end user can use Notepad to directly edit the file so there's no telling where these characters might embed themselves -- could be other than at the start or end of the string. So removing them all does it.

The other entries all work but Remove might be the easiest?

I learned quite a bit more about Regex from this post -- pretty cool work with its use here.

blocks|key|3990714|text|我用了很多擦除级别|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3990715|String+donen+=+"lots+of+stupid+whitespaces+and+new+lines+and+others..."

//Remove+multilines
donen+=+Regex.Replace(donen,+@"%5E\s%2B$[\r\n]*",+"",+RegexOptions.Multiline);
//remove+multi+whitespaces
RegexOptions+options+=+RegexOptions.None;
Regex+regex+=+new+Regex("[+]{2,}",+options);
donen+=+regex.Replace(donen,+"+");
//remove+tabs
char+tab+=+'\u0009';
donen+=+donen.Replace(tab.ToString(),+"");
//remove+endoffile+newlines
donen+=+donen.TrimEnd('\r',+'\n');
//to+be+sure+erase+new+lines+again+from+another+perspective
donen.Replace(Environment.NewLine,+"");|code-block|syntax|javascript|3990716|现在我们有了干净的一行|3990717|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|$]]

I use lots of erasing level

<pre><code>String donen = "lots of stupid whitespaces and new lines and others..."

//Remove multilines
donen = Regex.Replace(donen, @"^\s+$[\r\n]*", "", RegexOptions.Multiline);
//remove multi whitespaces
RegexOptions options = RegexOptions.None;
Regex regex = new Regex("[ ]{2,}", options);
donen = regex.Replace(donen, " ");
//remove tabs
char tab = '\u0009';
donen = donen.Replace(tab.ToString(), "");
//remove endoffile newlines
donen = donen.TrimEnd('\r', '\n');
//to be sure erase new lines again from another perspective
donen.Replace(Environment.NewLine, "");
</code></pre>

and now we have a clean one row

blocks|key|1252188|text|这就是我要为自己工作的东西。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1252189|s.Replace("\r","").Replace("\n","")|code-block|syntax|javascript|1252190|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|$]]

This is what I got to work for me.

<pre><code>s.Replace("\r","").Replace("\n","")
</code></pre>

blocks|key|4836071|text|varName.replace(/[\r\n]/mg,+'')++++++++++++++++++++++++++++++++++++++++++++++|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4836072|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>varName.replace(/[\r\n]/mg, '') 
</code></pre>

How do I remove the carriage return character <code>(\r)</code> and the new line character<code>(\n)</code> from the end of a string?

Removing carriage return and new-line from the end of a string in c#

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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如何删除字符串末尾的回车字符(\r)和换行符(\n)？

问在c#中从字符串末尾删除回车符和换行符
EN

回答 12

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