正则表达式选择引号之外的字符 [英] Regex to pick characters outside of pair of quotes

问题描述

我想找到一个正则表达式，它可以挑选出引号集之外的所有逗号.

I would like to find a regex that will pick out all commas that fall outside quote sets.

例如:

'foo' => 'bar',
'foofoo' => 'bar,bar'

这将选出第 1 行 'bar',

我不太关心单引号和双引号.

I don't really care about single vs double quotes.

有没有人有任何想法?我觉得这应该可以通过预读来实现，但是我的正则表达式 fu 太弱了.

Has anyone got any thoughts? I feel like this should be possible with readaheads, but my regex fu is too weak.

推荐答案

这将匹配任何字符串，直到并包括第一个未加引号的,".这就是你想要的吗?

This will match any string up to and including the first non-quoted ",". Is that what you are wanting?

/^([^"]|"[^"]*")*?(,)/

如果你想要所有这些(并且作为对那个说不可能的人的反例)你可以写:

If you want all of them (and as a counter-example to the guy who said it wasn't possible) you could write:

/(,)(?=(?:[^"]|"[^"]*")*$)/

将匹配所有这些.于是

which will match all of them. Thus

'test, a "comma,", bob, ",sam,",here'.gsub(/(,)(?=(?:[^"]|"[^"]*")*$)/,';')

用分号替换引号内的所有非逗号，并产生:

replaces all the commas not inside quotes with semicolons, and produces:

'test; a "comma,"; bob; ",sam,";here'

如果您需要它跨换行符，只需添加 m(多行)标志.

If you need it to work across line breaks just add the m (multiline) flag.

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