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，如下所示:a:1,b:2,c:3".我想为指定的键提取一个值，例如我会为a"得到1".我打算用这样的正则表达式来做:

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:)^,*'

但似乎 grep 不支持环视.(我什至不确定这个正则表达式是否正常工作.)还有其他方法吗?

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.-P 启用 PCRE 引擎.双引号是字符串插值所必需的，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:

• (?:^|,) - 字符串或逗号的开始
• $KEY - KEY 变量
• : - 冒号
• K - 匹配重置运算符，丢弃目前匹配的整个文本
• [^,]+ - 除了 ,
之外的 1+ 个字符

• (?:^|,) - 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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