正则表达式以匹配所有不在引号内的实例 [英] Regex to match all instances not inside quotes

问题描述

来自此问与答，我推断出在引号内匹配给定正则表达式 not 的所有实例是不可能的。也就是说，它不能匹配转义的引号（例如：应使用整个\ match\ ）。如果有我不知道的方法可以解决我的问题。

From this q/a, I deduced that matching all instances of a given regex not inside quotes, is impossible. That is, it can't match escaped quotes (ex: "this whole \"match\" should be taken"). If there is a way to do it that I don't know about, that would solve my problem.

但是，如果没有，我想知道是否有任何有效的选择可以在JavaScript中使用。我已经考虑了一下，但是无法提供任何适用于大多数（即使不是全部）情况的优雅解决方案。

If not, however, I'd like to know if there is any efficient alternative that could be used in JavaScript. I've thought about it a bit, but can't come with any elegant solutions that would work in most, if not all, cases.

具体地说，我只需要替代方法即可工作使用.split（）和.replace（）方法，但是如果可以更通用的话，那将是最好的方法。

Specifically, I just need the alternative to work with .split() and .replace() methods, but if it could be more generalized, that would be the best.

例如：

输入字符串：
+ bar + baz not + or\ +或+ \ this + foo + bar +

用＃代替+，而不用引号引起来，将返回：
＃bar＃baz not + or\ + or + \; this +" foo＃bar＃

推荐答案

实际上，您可以匹配正则表达式的所有实例不是在任何字符串的引号内，每个引号都被再次关闭。像上面的示例一样，您要匹配 \ + 。

Actually, you can match all instances of a regex not inside quotes for any string, where each opening quote is closed again. Say, as in you example above, you want to match \+.

这里的关键观察是，如果单词后面有双引号，则该单词在引号外。可以将其建模为先行断言：

The key observation here is, that a word is outside quotes if there are an even number of quotes following it. This can be modeled as a look-ahead assertion:

\+(?=([^"]*"[^"]*")*[^"]*$)

现在，您不想计算转义的引号，这会变得更加复杂。代替前进到下一个引号的 [^] * 一样，您还需要考虑反斜杠并使用 [^ \\] * 。到达反斜杠或引号后，如果遇到反斜杠，则需要忽略下一个字符，否则前进到下一个未转义的引号。看起来像（\\。|（[[^ \\] * \\。）* [^ \\] *）。组合起来，您会到达

Now, you'd like to not count escaped quotes. This gets a little more complicated. Instead of [^"]* , which advanced to the next quote, you need to consider backslashes as well and use [^"\\]*. After you arrive at either a backslash or a quote, you need to ignore the next character if you encounter a backslash, or else advance to the next unescaped quote. That looks like (\\.|"([^"\\]*\\.)*[^"\\]*"). Combined, you arrive at

\+(?=([^"\\]*(\\.|"([^"\\]*\\.)*[^"\\]*"))*[^"]*$)

我承认这是一个小隐喻。=）

I admit it is a little cryptic. =)

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