删除括号前后的所有空格 [英] Remove all spaces before and after parentheses

问题描述

我想在括号前 和 之后删除一个或多个空格.跟随这篇文章，其中使用以下正则表达式解决了PHP的问题

I would like to remove one or more spaces before and after any parentheses. Following this post where the issue has been solved for PHP with the following regex

(?<=[([]) +| +(?=[)\]])

现在，我想在 Javascript 中执行相同的操作，但是Javascript正则表达式引擎的前瞻性和后瞻性与PHP相同.我设法使以下正则表达式至少在Javascript中有效，但是它删除了所有空格:

now I would like to do the same in Javascript but Javascript regex engine does not have the same lookahead and lookbehind as PHP. I managed to make the following regex at least work in Javascript but it removes all spaces:

?![([]) +| +(?=[)\]])

请参见 Regex101测试.

给出字符串:

This is ( a sample     ) [           string ] to play with

预期结果:

This is (a sample) [string] to play with

推荐答案

您可以改用捕获组，并用其占位符替换以恢复结果中的括号/括号:

You may use capturing groups instead and replace with their placeholders to restore the bracket/parentheses in the result:

.replace(/([([])\s+|\s+([)\]])/g, "$1$2");

请参见 regex演示

详细信息

• ([[[[])\ s + -组1捕获(或 [(用 $ 1引用(来自字符串替换模式)，然后是1+空格
• | -或
• \ s +([] \]])-1个以上的空格，后跟一个)或] 捕获到第2组中(称为字符串替换模式中的 $ 2 )

• ([([])\s+ - Group 1 capturing either a ( or a [ (referred to with $1 from the string replacement pattern) and then 1+ whitespaces
• | - or
• \s+([)\]]) - 1+ whitespaces followed with a ) or ] captured into Group 2 (referred to with $2 from the string replacement pattern)

JS演示:

var strs = ['This is ( a sample ) [ string ] to play with', 
            'This is ( a sample     ) [           string] to play with'];
var rx = /([([])\s+|\s+([)\]])/g;
for (var s of strs) {
  console.log(s+ " =>\n"+ s.replace(rx, "$1$2"));
}

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