[英]JavaScript: how to use a regular expression to remove blank lines from a string?
I need to use JavaScript to remove blank lines in a HTML text box.我需要使用 JavaScript 删除 HTML 文本框中的空行。 The blank lines can be at anywhere in the
textarea
element.空行可以位于
textarea
元素中的任何位置。 A blank line can be just a return or white spaces plus return.空行可以只是一个回车或空格加回车。
I am expecting a regular expression solution to this.我期待正则表达式解决方案。 Here are some I tried, but they are not working and cannot figure out why:
以下是我尝试过的一些方法,但它们不起作用,无法弄清楚原因:
/^\s*\r?\n/g
/^\s*\r?\n$/g
Edit 1编辑 1
It appears that the solution (I modified it a little) suggested by aaronman and m.buettner works:看来 aaronman 和 m.buettner 建议的解决方案(我稍微修改了一下)有效:
string.replace(/^\s*\n/gm, "")
Can someone tell why my first regular expression is not working?有人能说出为什么我的第一个正则表达式不起作用吗?
Edit 2编辑 2
After reading all useful answers, I came up with this:在阅读了所有有用的答案后,我想出了这个:
/^[\s\t]*(\r\n|\n|\r)/gm
Is this going to be one that cover all situations?这将是一种涵盖所有情况的方法吗?
Edit 3编辑 3
This is the most concise one covering all spaces (white spaces, tabs) and platforms (Linux, Windows, Mac).这是涵盖所有空间(空格、制表符)和平台(Linux、Windows、Mac)的最简洁的一种。
/^\s*[\r\n]/gm
Many thanks to m.buettner!非常感谢 m.buettner!
Your pattern seems alright, you just need to include the multiline modifier m
, so that ^
and $
match line beginnings and endings as well:您的模式似乎没问题,您只需要包含多行修饰符
m
,以便^
和$
匹配行的开头和结尾:
/^\s*\n/gm
Without the m
, the anchors only match string-beginnings and endings.没有
m
,锚只匹配字符串的开头和结尾。
Note that you miss out on UNIX-style line endings (only \\r
).请注意,您错过了 UNIX 样式的行尾(仅
\\r
)。 This would help in that case:在这种情况下,这将有所帮助:
/^\s*[\r\n]/gm
Also note that (in both cases) you don't need to match the optional \\r
in front of the \\n
explicitly, because that is taken care of by \\s*
.另请注意,(在这两种情况下)您不需要显式匹配
\\n
前面的可选\\r
,因为这是由\\s*
。
As Dex pointed out in a comment, this will fail to clear the last line if it consists only of spaces (and there is no newline after it).正如Dex在评论中指出的那样,如果最后一行仅包含空格(并且后面没有换行符),则它无法清除最后一行。 A way to fix that would be to make the actual newline optional but include an end-of-line anchor before it.
解决这个问题的一种方法是使实际的换行符可选,但在它之前包含一个行尾锚点。 In this case you do have to match the line ending properly though:
在这种情况下,你必须符合正确的结束,虽然该行:
/^\s*$(?:\r\n?|\n)/gm
我相信这会奏效
searchText.replace(/(^[ \t]*\n)/gm, "")
This should do the trick i think:这应该可以解决我认为的问题:
var el = document.getElementsByName("nameOfTextBox")[0];
el.value.replace(/(\r\n|\n|\r)/gm, "");
EDIT: Removes three types of line breaks.编辑:删除三种类型的换行符。
Here's another solution :这是另一个解决方案:
string = string.replace(/^(?=\n)$|^\s*|\s*$|\n\n+/gm, "")
It appears to handle both blank lines and space-only lines.它似乎处理空行和仅空格行。
function removeEmptyLine(text) {
return text.replace(/(\r?\n)\s*\1+/g, '$1');
}
test:测试:
console.assert(removeEmptyLine('a\r\nb') === 'a\r\nb');
console.assert(removeEmptyLine('a\r\n\r\nb') === 'a\r\nb');
console.assert(removeEmptyLine('a\r\n \r\nb') === 'a\r\nb');
console.assert(removeEmptyLine('a\r\n \r\n \r\nb') === 'a\r\nb');
console.assert(removeEmptyLine('a\r\n \r\n 2\r\n \r\nb') === 'a\r\n 2\r\nb');
console.assert(removeEmptyLine('a\nb') === 'a\nb');
console.assert(removeEmptyLine('a\n\nb') === 'a\nb');
console.assert(removeEmptyLine('a\n \nb') === 'a\nb');
console.assert(removeEmptyLine('a\n \n \nb') === 'a\nb');
console.assert(removeEmptyLine('a\n \n2 \n \nb') === 'a\n2 \nb');
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