正则表达式删除除括号之间的所有空格 [英] regex to remove all whitespaces except between brackets

问题描述

我一直在努力解决一个我希望用正则表达式解决的问题.

I've been wrestling with an issue I was hoping to solve with regex.

假设我有一个字符串，它可以包含任何字母数字，并且可能有一个子字符串被方括号包围.这些子字符串可以像这样出现在字符串中的任何位置.也可以有任意数量的括号括起来的子串.

Let's say I have a string that can contain any alphanumeric with the possibility of a substring within being surrounded by square brackets. These substrings could appear anywhere in the string like this. There can also be any number of bracket-ed substrings.

示例:

• aaa[bb b]
• aaa[bbb]ccc[d dd]
• [aaa]bbb[c cc]

你可以看到一些括号中的子串中有空格，这很好.我现在的主要问题是当我遇到括号外的空格时:

You can see that there are whitespaces in some of the bracketed substrings, that's fine. My main issue right now is when I encounter spaces outside of the brackets like this:

• a aa[bb b]

现在我想保留括号内的空格，但在其他地方删除它们.

Now I want to preserve the spaces inside the brackets but remove them everywhere else.

对于像这样的字符串，这会变得有点棘手:

This gets a little more tricky for strings like:

• a aa[bb b]c cc[d dd]e ee[f ff]

这里我希望返回:

• aaa[bb b]ccc[d dd]eee[f ff]

我现在花了一些时间阅读有关环视、否定断言等的不同 reg ex 页面，这让我头晕目眩.

I spent some time now reading through different reg ex pages regarding lookarounds, negative assertions, etc. and it's making my head spin.

注意:对于访问这里的任何人，我并不是在寻找任何涉及嵌套括号的解决方案.如果是这样的话，我可能会像下面提到的一些评论一样务实地去做.

NOTE: for anyone visiting this, I was not looking for any solution involving nested brackets. If that was the case I'd probably do it pragmatically like some of the comments mentioned below.

推荐答案

这个正则表达式应该可以解决问题:

This regex should do the trick:

[ ](?=[^\]]*?(?:\[|$))

只需替换与"匹配的空格即可.

Just replace the space that was matched with "".

基本上它所做的就是确保您要删除的空间前面有一个["，但如果它前面有一个]"，则不会.

Basically all it's doing is making sure that the space you are going to remove has a "[" in front of it, but not if it has a "]" before it.

只要您没有嵌套的方括号，这应该可以工作，例如:

That should work as long as you don't have nested square brackets, e.g.:

a a[b [c c]b]

a a[b [c c]b]

因为在这种情况下，第一个b"后面的空格将被删除，它会变成:

Because in that case, the space after the first "b" will be removed and it will become:

aa[b[c c]b]

aa[b[c c]b]

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