按空格拆分字符串，保留带引号的段，允许转义引号 [英] Split a string by whitespace, keeping quoted segments, allowing escaped quotes

问题描述

我目前有这个正则表达式来按所有空格分割字符串，除非它在引用的段中：

I currently have this regular expression to split strings by all whitespace, unless it's in a quoted segment:

keywords = 'pop rock "hard rock"';
keywords = keywords.match(/\w+|"[^"]+"/g);
console.log(keywords); // [pop, rock, "hard rock"]

但是，我也希望它可以在关键字中引用，例如：

However, I also want it to be possible to have quotes in keywords, like this:

keywords = 'pop rock "hard rock" "\"dream\" pop"';

这应该返回

[pop, rock, "hard rock", "\"dream\" pop"]

什么是最简单的实现这一目标的方法？

What's the easiest way to achieve this?

推荐答案

您可以将正则表达式更改为：

You can change your regex to:

keywords = keywords.match(/\w+|"(?:\\"|[^"])+"/g);

而不是 [^] + 你有（？：\\| [^]）+ 允许 \或其他角色，但不是未转义的引用。

Instead of [^"]+ you've got (?:\\"|[^"])+ which allows \" or other character, but not an unescaped quote.

一个重要的注意事项是，如果你想让字符串包含一个文字斜杠，它应该是b e：

One important note is that if you want the string to include a literal slash, it should be:

keywords = 'pop rock "hard rock" "\\"dream\\" pop"'; //note the escaped slashes.

此外， \w + 和 [^] + - 例如，它将匹配单词ab * d，但不是 ab * d （没有引号）。考虑使用 [^\ s] + 代替，这将匹配非空格。

Also, there's a slight inconsistency between \w+ and [^"]+ - for example, it will match the word "ab*d", but not ab*d (without quotes). Consider using [^"\s]+ instead, that will match non-spaces.

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