替代正则表达式：匹配所有不在引号内的实例 [英] Alternative to regex: match all instances not inside quotes

问题描述

从这个q / a ，我推断，匹配引号内的给定正则表达式 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：这整个\match\应该被采用）。如果有办法做到这一点我不了解，那就解决我的问题。

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.

然而，如果没有，我想知道是否有是可以在JavaScript中使用的任何有效的替代方案。我已经想了一下，但是不能在大多数（如果不是全部）情况下都可以使用任何优雅的解决方案。

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.

具体来说，我只需要使用.split（）和.replace（）方法的替代方法，但如果可以更广泛地使用，那将是最好的。

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.

：

输入字符串：
+ bar + baznot + or\+或+ \this +foo + bar + / code>

替换+与＃，而不是内部的引号，将返回：
＃bar＃baznot + or\+或+ \\this +foo＃bar＃

推荐答案

实际上，你可以匹配正则表达式的所有实例没有任何字符串的引号，每个开头报价再次关闭。说，如上例所示，你想要匹配 \ + 。

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，一个字是外面的引号，如果有一个偶数的引号跟随它。这可以被建模为一个先行断言：

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，你不想计数转义的引号，这会变得更复杂一些，而不是 [^] * ，它提前到下一个引用，您还需要考虑反斜杠，并使用 [^\\] * 。在您收到反斜杠或引号后，您需要忽略下一个字符，如果您遇到反斜杠，或者进入看起来像（\\。。）（[^\\] * \\。）* [^\\] *）。结合起来，你到达

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 cryptic。=）

I admit it is a little cryptic. =)

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