使用grep从键值对列表中提取值 [英] Extract value from a list of key-value pairs using grep

问题描述

我有一个包含键值对列表的字符串，例如:"a:1，b:2，c:3".我想提取指定键的值，例如我将为"a"得到"1".我打算用这样的正则表达式来做到这一点:

I have a string containing a list of key-value pairs like this: "a:1,b:2,c:3". I would like to extract a value for a specified key so that e.g. I would get "1" for "a". I was planning to do it with a regex like this:

'(?<=(^|,)$KEY:)^,*'

，但是grep似乎不支持环视. (我什至不确定此正则表达式是否正常工作.)还有另一种方法吗?

but it seems grep doesn't support lookarounds. (I'm not even sure this regex works correctly.) Is there another way?

推荐答案

您可以使用

grep -oP "(?:^|,)$KEY:\K[^,]+"

-o选项输出匹配项. -P启用PCRE引擎.字符串插值必须使用双引号，以便可以对$KEY进行插值.

The -o option outputs matches. -P enables PCRE engine. The double quotes are necessary for string interpolation so that $KEY could be interpolated.

模式匹配:

• (?:^|,)-字符串或逗号开头
• $KEY-KEY变量
• :-冒号
• \K-匹配重置运算符，该运算符将舍弃到目前为止已匹配的整个文本
• [^,]+-除,
以外的1个以上字符

• (?:^|,) - start of string or comma
• $KEY - the KEY variable
• : - colon
• \K - match reset operator that discards the whole text matched so far
• [^,]+ - 1+ chars other than ,

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