sed JSON正则表达式 [英] sed JSON regular expression

问题描述

为这样的假问题道歉，但这是我第一次使用curl命令，现在我从某个地方得到了该命令以提取以下字符串

Apologize for such dummy question but this is my first time using curl command and now I got this command from somewhere to extract the following string

{成功":true，结果":1，总计":1，更多":false，偏移量":0，命中":[{路径":"/home/users/Vq7DPVRHzGVK--OTJsHs，"摘录:"，"名称:" Vq7DPVRHzGVK--OTJsHs，"标题:" Vq7DPVRHzGVK--OTJsHs，" lastModified:" 2017-03-03 16:45:46，"创建:" 2017-03-03 16:45:46}]}

{"success":true,"results":1,"total":1,"more":false,"offset":0,"hits":[{"path":"/home/users/Vq7DPVRHzGVK--OTJsHs","excerpt":"","name":"Vq7DPVRHzGVK--OTJsHs","title":"Vq7DPVRHzGVK--OTJsHs","lastModified":"2017-03-03 16:45:46","created":"2017-03-03 16:45:46"}]}

我使用以下脚本将curl输出传送到sed:

I pipe the curl output to sed with the following script:

sed -e 's/^.*"path":"\([^"]*\)".*$/\1/

结果:

/home/users/Vq7DPVRHzGVK--OTJsHs

/home/users/Vq7DPVRHzGVK--OTJsHs

任何人都可以解释正则表达式在这里的工作方式吗?以及如何仅获得 Vq7DPVRHzGVK--OTJsHs 的结果，而不包括/home/user路径?

Can anyone explain how's the regex work here? and how do I get the result for only Vq7DPVRHzGVK--OTJsHs instead of including the /home/user path?

推荐答案

说明:

s/   ^.*"path":"\([^"]*\)".*$   /  \1   /
      ----------^------------     ---^---
             Pattern         Replacement string 

正则表达式如何工作:

^.*         # Match beginning of input string & anything else
"path":"    # Up to literal string `"path":"`
\([^"]*\)   # Then match slash and match + group anything up to a double quote `"`
".*$        # Match double quote and the rest of input string

通过替换字符串\1，您正在将整个匹配部分替换为第一个捕获组，该捕获组是路径值双引号之间的所有内容，除了开始斜杠之外.

By replacement string \1 you are replacing whole matching part with first capturing group which is every thing between double quotes of path value except the beginning slash.

您想要的是将捕获组从捕获整个部分更改为最后一部分:

What you want is changing capturing group from capturing whole part to last section:

s/^.*"path":"[^"]*\/\([^"]*\)".*$/\1/

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