[英]Regex lazy on non-capturing group
I have this regex: 我有这个正则表达式:
(?:(?:AND\sNOT|AND|OR)(?!.*(?:AND\sNOT|AND|OR))\s)(.*)
What I want is to get the last key:value pair, example - 我想要的是得到最后一个关键:值对,例子 -
k:v AND k1:v1 AND NOT k2:v2 OR k3:v3
I want the regex to match k3:v3, and it does, but it doesn't match the key:value in the next situation - 我希望正则表达式匹配k3:v3,它确实匹配,但它与下一个情境中的键:值不匹配 -
k:v
I need it to match the key:value pair even if it's the first one and there's no operators brefore... 我需要它来匹配键:值对,即使它是第一个并且没有运算符brefore ...
update: key:value is not the issue here, I need it to match everything after the last operator 更新:键:值不是这里的问题,我需要它匹配最后一个运算符后的所有内容
update 2: tried to do the following - 更新2:尝试执行以下操作 -
(?:(?:AND\sNOT|AND|OR)?(?!.*(?:AND\sNOT|AND|OR))\s)(.*)
(?:(?:AND\sNOT|AND|OR?)(?!.*(?:AND\sNOT|AND|OR))\s)(.*)
didn't work 没用
Edit - My bad, \\K
isn't supported in JS. 编辑 - 我的坏,在JS中不支持
\\K
Here's an alternative one: 这是另一种选择:
/(?:.*(?:AND|OR|NOT)\\s+)?(.*)/
(group 1) /(?:.*(?:AND|OR|NOT)\\s+)?(.*)/
)
https://regex101.com/r/AyVV89/3 (Please check the matches on the right panel, for some reason group 1 isn't highlighted on the text.) https://regex101.com/r/AyVV89/3 (请检查右侧面板上的匹配项,由于某种原因,文本中未突出显示组1。)
Original post 原帖
Per your last update (that wasn't obvious at all before you mentioned it): 根据您的上次更新(在您提及之前根本不明显):
key:value is not the issue here, I need it to match everything after the last operator
key:value不是问题,我需要它匹配最后一个运算符后的所有内容
This one will match anything after the last operator, no matter what it is (and work without an operator too): 这个将匹配最后一个运算符之后的任何内容,无论它是什么(并且没有运算符也可以):
/(?:.*(?:AND|OR|NOT)\\s+\\K)?.*/
https://regex101.com/r/AyVV89/2 https://regex101.com/r/AyVV89/2
Update 更新
As per comments, I assume you need to match last key:value pair along with its operator if any, and anything that comes after: 根据评论,我假设您需要匹配最后一个键:值对及其运算符(如果有)以及之后的任何内容:
(?:.*\b(AND(?:\sNOT)?|OR) +)?(\S+:\S+.*)
What you need could be simplified into this: 你需要什么可以简化为:
(?:AND(?:\sNOT)?|OR)\s+(\S+)$
(?:AND(?:\\sNOT)?|OR)
match AND
or AND NOT
or OR
(?:AND(?:\\sNOT)?|OR)
匹配AND
或AND NOT
或OR
\\s+
match any number of whitespaces \\s+
匹配任意数量的空格 (\\S+)
match and capture none-whitespace characters (\\S+)
匹配并捕获无空白字符 $
asserts end of string $
asserts字符串结尾 From your explanation, you don't need the first operator at all. 根据您的解释,您根本不需要第一个操作员。 Simply try
试试吧
(?!.*(?:NOT|AND|OR)\s)(\b.*)
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