blocks|key|283058|text|arrayOfLines+=+lineString.match(/[%5E\r\n]%2B/g);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|283059|正如Tim所说，这既是整个匹配，也是捕获。无论全局修饰符是什么，regex.exec(string)似乎都会在找到第一个匹配项时返回，而string.match(regex)则是全局匹配。|unstyled|offset|length|style|CODE|283060|entityMap^0|0|W|I|1W|J|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>arrayOfLines = lineString.match(/[^\r\n]+/g);
</code></pre>

As Tim said, it is both the entire match and capture. It appears <code>regex.exec(string)</code> returns on finding the first match regardless of global modifier, wheras <code>string.match(regex)</code> is honouring global.

blocks|key|277437|text|使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|277438|result+=+subject.split(/\r?\n/);|code-block|syntax|javascript|277439|正则表达式返回line1两次，因为line1既是整个匹配，也是第一个捕获组的内容。|offset|length|style|CODE|277440|entityMap^0|0|0|7|5|H|5|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]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

Use

<pre><code>result = subject.split(/\r?\n/);
</code></pre>

Your regex returns <code>line1</code> twice because <code>line1</code> is both the entire match and the contents of the first capturing group.

blocks|key|277453|text|我假设下面是换行符|type|unstyled|depth|inlineStyleRanges|entityRanges|data|277454|277455|\r依次为\n+|ordered-list-item|277456|\n和\r+|277457|\n单独存在|277458|\r单独存在|277459|277460|请使用|277461|var+re=/\r\n%7C\n\r%7C\n%7C\r/g;

arrayofLines=lineString.replace(re,"\n").split("\n");|code-block|syntax|javascript|277462|获取所有行的数组，包括空行。|277463|或|offset|length|style|BOLD|277464|277465|arrayOfLines+=+lineString.match(/[%5E\r\n]%2B/g);+|277466|对于非空行的数组|277467|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|19|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|1A|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|1B|8|@]|9|@]|A|$]]|$1|H|3|I|5|E|7|1C|8|@]|9|@]|A|$]]|$1|J|3|K|5|E|7|1D|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1F|8|@]|9|@]|A|$]]|$1|O|3|P|5|Q|7|1G|8|@]|9|@]|A|$R|S]]|$1|T|3|U|5|6|7|1H|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|1I|8|@$X|1J|Y|1K|Z|10]]|9|@]|A|$]]|$1|11|3|N|5|6|7|1L|8|@]|9|@]|A|$]]|$1|12|3|13|5|Q|7|1M|8|@]|9|@]|A|$R|S]]|$1|14|3|15|5|6|7|1N|8|@]|9|@]|A|$]]|$1|16|3|-4|5|6|7|1O|8|@]|9|@]|A|$]]]|17|$]]

I am assuming following constitute newlines

<ol>
<li>\r followed by \n </li>
<li>\n followed by \r </li>
<li>\n present alone </li>
<li>\r present alone</li>
</ol>

Please Use

<pre><code>var re=/\r\n|\n\r|\n|\r/g;

arrayofLines=lineString.replace(re,"\n").split("\n");
</code></pre>

for an array of all Lines including the empty ones.

OR

Please Use

<pre><code>arrayOfLines = lineString.match(/[^\r\n]+/g); 
</code></pre>

For an array of non empty Lines

blocks|key|283082|text|更简单的正则表达式处理所有行结束组合，甚至混合在同一文件中，并删除空行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|283083|var+lines+=+text.split(/[\r\n]%2B/g);|offset|length|style|CODE|283084|使用空格修剪：|283085|var+lines+=+text.trim().split(/\s*[\r\n]%2B\s*/g);|283086|entityMap^0|0|0|Z|0|0|0|1C|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@$D|P|E|Q|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|R|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|S|8|@$D|T|E|U|F|G]]|9|@]|A|$]]|$1|L|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|M|$]]

Even simpler regex that handles all line ending combinations, even mixed in the same file, and removes empty lines as well:

<code>var lines = text.split(/[\r\n]+/g);</code>

With whitespace trimming:

<code>var lines = text.trim().split(/\s*[\r\n]+\s*/g);</code>

blocks|key|279217|text|首先将所有\r\n替换为\n，然后是String.split。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|279218|entityMap^0|5|4|C|2|I|C|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

First replace all <code>\r\n</code> with <code>\n</code>, then <code>String.split</code>.

blocks|key|283088|text|符合Unicode的行拆分|type|unstyled|depth|inlineStyleRanges|entityRanges|data|283089|Unicode®技术标准#18定义了line+boundaries的构成。同一部分还提供了一个正则表达式来匹配所有行边界。使用该正则表达式，我们可以定义以下JS函数，该函数在任意行边界处拆分给定的字符串(保留空行以及前导和尾随空格)：|offset|length|283090|const+splitLines+=+s+=>+s.split(/\r\n%7C(?!\r\n)[\n-\r\x85\u2028\u2029]/)|code-block|syntax|javascript|283091|我不明白为什么需要负向前瞻部分((?!\r\n))，但这就是Unicode文档‍♂️中所建议的。|style|CODE|283092|上面的文档建议定义一个正则表达式元字符，用于匹配所有行结束字符和序列。Perl+has+\R就是这样的。不幸的是，JavaScript没有包含这样的元字符。遗憾的是，我甚至找不到TC39对此的提议。|283093|entityMap|0|LINK|mutability|MUTABLE|url|http://www.unicode.org/reports/tr18/#Line_Boundaries|1|https://perldoc.perl.org/perlrebackslash#%255CR^0|0|I|F|0|0|0|G|8|0|18|2|Z|B|1|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@$D|12|E|13|1|14]]|A|$]]|$1|F|3|G|5|H|7|15|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|16|8|@$D|17|E|18|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|19|8|@$D|1A|E|1B|M|N]]|9|@$D|1C|E|1D|1|1E]]|A|$]]|$1|Q|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]|Y|$5|T|U|V|A|$W|Z]]]]

<h3>Unicode Compliant Line Splitting</h3>
Unicode® Technical Standard #18 defines what constitutes <a href="http://www.unicode.org/reports/tr18/#Line_Boundaries" rel="nofollow noreferrer">line boundaries</a>. That same section also gives a regular expression to match all line boundaries. Using that regex, we can define the following JS function that splits a given string at any line boundary (preserving empty lines as well as leading and trailing whitespace):
<pre class="lang-js prettyprint-override"><code>const splitLines = s =&gt; s.split(/\r\n|(?!\r\n)[\n-\r\x85\u2028\u2029]/)
</code></pre>
I don't understand why the negative look-ahead part (<code>(?!\r\n)</code>) is necessary, but that is what is suggested in the Unicode document ‍♂️.
The above document recommends to define a regular expression meta-character for matching all line ending characters and sequences. <a href="https://perldoc.perl.org/perlrebackslash#%5CR" rel="nofollow noreferrer">Perl has <code>\R</code></a> for that. Unfortunately, JavaScript does not include such a meta-character. Alas, I could not even find a TC39 proposal for that.

blocks|key|281109|text|http://jsfiddle.net/uq55en5o/|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|281110|var+lines+=+text.match(/%5E.*((\r\n%7C\n%7C\r)%7C$)/gm);|style|CODE|281111|我曾经做过这样的事情。上面的链接是我的小提琴。|281112|entityMap|0|LINK|mutability|MUTABLE|url^0|0|T|0|0|0|1C|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|6|7|U|8|@$A|V|B|W|F|G]]|9|@]|C|$]]|$1|H|3|I|5|6|7|X|8|@]|9|@]|C|$]]|$1|J|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|K|$L|$5|M|N|O|C|$P|4]]]]

<a href="http://jsfiddle.net/uq55en5o/" rel="nofollow">http://jsfiddle.net/uq55en5o/</a>

<code>var lines = text.match(/^.*((\r\n|\n|\r)|$)/gm);</code>

I have done something like this. Above link is my fiddle.

How do you split a long piece of text into separate lines? Why does this return line1 twice?

<pre><code>/^(.*?)$/mg.exec('line1\r\nline2\r\n');
</code></pre>

<blockquote>
 ["line1", "line1"]
</blockquote>

I turned on the multi-line modifier to make <code>^</code> and <code>$</code> match beginning and end of lines. I also turned on the global modifier to capture all lines.

I wish to use a regex split and not <code>String.split</code> because I'll be dealing with both Linux <code>\n</code> and Windows <code>\r\n</code> line endings.

JS regex to split by line

Windows

Linux

如何将一段很长的文本拆分成单独的行？为什么这两次返回line1？/^(.*?)$/mg.exec('line1\r\nline2\r\n'); "line1"，"line1“我打开了多行修饰符，使^和$匹配行首和行尾。我还打开了全局修改器来捕获所有行。我希望使用正则表达式拆分，而不是String.split，因为我将同时处理Linux \n和Windows \r\n行结尾。

问JS正则表达式按行拆分
EN

回答 7

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问JS正则表达式按行拆分EN

回答 7

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问JS正则表达式按行拆分
EN