Javascript正则表达式适用于不在某些字符之间的所有单词 [英] Javascript Regex for all words not between certain characters

问题描述

我正在尝试返回不包含在方括号之间的所有单词的计数.所以给定..

I'm trying to return a count of all words NOT between square brackets. So given ..

[don't match these words] but do match these

最后四个单词的计数为4.

I get a count of 4 for the last four words.

这在.net中有效:

\b(?<!\[)[\w']+(?!\])\b

但是它在Javascript中不起作用，因为它不支持向后看

but it won't work in Javascript because it doesn't support lookbehind

对纯js正则表达式解决方案有什么想法吗?

Any ideas for a pure js regex solution?

推荐答案

好的，我认为这应该可行:

Ok, I think this should work:

\[[^\]]+\](?:^|\s)([\w']+)(?!\])\b|(?:^|\s)([\w']+)(?!\])\b

您可以在这里进行测试:
http://regexpal.com/

You can test it here:
http://regexpal.com/

如果您需要一个替代方案，并且要在主要文本之后加上方括号，则可以将其添加为第二替代方案，而当前的第二替代方案将成为第三替代方案.
有点复杂，但我现在想不出更好的解决方案.

If you need an alternative with text in square brackets coming after the main text, it could be added as a second alternative and the current second one would become third.
It's a bit complicated but I can't think of a better solution right now.

如果您需要对实际匹配项进行操作，则会在捕获组中找到它们.

If you need to do something with the actual matches you will find them in the capturing groups.

更新:

说明:因此，我们在这里有两个选择:

Explanation: So, we've got two options here:

1. \ [[[^ \]] + \](?:^ | \ s)([\ w'] +)(?！\])\ b

这是说:

• \ [[[^ \]] + \] -匹配方括号中的所有内容(不捕获)
• (?:^ | \ s)-后跟行首或空格-当我查看它时，请删除插入符号，因为它没有意义，所以这将变成<代码> \ s
• ([[ww]] +)-匹配以下所有单词字符，只要(?！\])下一个字符不是右括号-很好现在可能也没有必要了，所以让我们尝试删除前瞻
• \ b -并匹配单词边界

• \[[^\]]+\] - match everything in square brackets (don't capture)
• (?:^|\s) - followed by line start or a space - when I look at it now take the caret out as it doesn't make sense so this will become just \s
• ([\w']+) - match all following word characters as long as (?!\])the next character is not the closing bracket - well this is probably also unnecessary now, so let's try and remove the lookahead
• \b - and match word boundary

2 (?:^ | \ s)([\ w'] +)(?！\])\ b

如果找不到选项1，则只进行单词匹配，而不要查找方括号，因为我们在第一部分中确保方括号不在此处.

If you cannot find the option 1 - do just the word matching, without looking for square brackets as we ensured with the first part that they are not here.

好吧，所以我删除了所有不需要的东西(它们留在那里，因为在它起作用之前我尝试了很多选择:-)，而修改后的正则表达式如下:

Ok, so I removed all the things that we don't need (they stayed there because I tried quite a few options before it worked:-) and the revised regex is the one below:

\[[^\]]+\]\s([\w']+)(?!\])\b|(?:^|\s)([\w']+)\b

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