正则不严格遵循负面的lookahead？
我有一条正则是与没有引号（双或单个）的URL匹配的 and and amp; （ampersand）在url 中。
我做的正则是我做的
([^'` "\n]+)\.([^ \n]+)&([^ "`'\n]+)(?!["'])
，但这只是不拿最后一句话，而是匹配url
https://regex101.com/ r/vpmqzh/1
以上面图的示例为
google.com/cool?cool1=yes&Cool2=no&Amp;Cool3=no“
URL最终不应匹配”，
但它只是不匹配'o ' 并匹配剩余的URL。
我要做的是，如果最终存在此双重报价，那么只是不匹配整个URL。
I have a regex which is matching urls which don't have quotes (double or single) at the end and have & (ampersand) in the url.
The regex i made
([^'` "\n]+)\.([^ \n]+)&([^ "`'\n]+)(?!["'])
but it's just not taking the last word and matching the url
https://regex101.com/r/vpmqZH/1
Take the example of picture above
google.com/cool?cool1=yes&cool2=no&cool3=no"
the url should not match as it have " in the end
but it's just not matching 'o'
and matching the remaining url.
All I wanted to do is if this double quote is present in the end then just don't match the whole url.
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评论（2）
您需要在整个部分中使lookahead活跃。然后，我们可以选择
$行的结尾或
（？= \ s）空间的正lookahead。参见 https://regex101.com/r/6zgpsx/1
You need to make the lookahead active on the whole part after the ampersand. We then have the option of
$end of the lineor
(?=\s)positive lookahead for a space.See https://regex101.com/r/6ZGpSX/1
仅对于匹配，您可以省略捕获组，并使用否定的字符类，如果要允许匹配，则应从被否定的字符类中省略
`。说明
[^'“ \ s。]+匹配1+以外的非whitespace chars”'。'＆amp;＆amp;字面上匹配[^\ s“'']+匹配1+ 1+非whitespace chars除外，“'（？！\ s）断言右侧的空间边界，请参阅a REGEX DEMO 。
For a match only, you can omit the capture groups, and use a negated character class and you should omit the backtick
`from the negated character class if you want to allow to match it.Explanation
[^'"\s.]+Match 1+ non whitespace chars other than"'.\.Match a dot[^\s'"&]+Match 1+ non whitespace chars other than"'&&Match literally[^\s"']+Match 1+ non whitespace chars other than"'(?!\S)Assert a whitespace boundary to the rightSee a regex demo.