grep正则表达式在两个已知字符串之间拉出一个字符串 [英] grep regex to pull out a string between two known strings

问题描述

我在解析出的一个文件中有一串文本，我几乎搞清楚了，但不知道我在想什么

<我使用的基本表达式是

cat cred.txt | grep -m 1 -o'& CD = [^&] *'


$ b

 & CD = u8AA-RaF-97gc_SdZ0J74gc_SdZ0J196gc_SdZ0J211 
  

我不希望& CD = 部分 strong>在结果字符串中，我该怎么做。

我解析的字符串是：

  webpage.asp？UserName = username& CD = u8AA-RaF-97gc_SdZ0J74gc_SdZ0J196gc_SdZ0J211& Country = USA 
   grep 知道Perl正则表达式：
 $ p> 
 
解决方案
 b 
 $ b 
  grep -m 1 -oP'（？< =& CD =）[^&] *'cred.txt 
  
如果不是： 
 
 
  sed'1s /.*& amp; CD = \（[^&] * \）。* / \ 1 /'cred.txt 
  
 


I have a string of text in a file that I am parsing out, I almost got it but not sure what I am missing


basic expression I am using is
cat cred.txt | grep -m 1 -o '&CD=[^&]*'
I am getting a results of 
&CD=u8AA-RaF-97gc_SdZ0J74gc_SdZ0J196gc_SdZ0J211
I do not want the &CD= part in the resulting string, how would I do that.

The string I am parsing from is:
webpage.asp?UserName=username&CD=u8AA-RaF-97gc_SdZ0J74gc_SdZ0J196gc_SdZ0J211&Country=USA

解决方案
If your grep knows Perl regex:
grep -m 1 -oP '(?<=&CD=)[^&]*' cred.txt
If not:
sed '1s/.*&CD=\([^&]*\).*/\1/' cred.txt


                        
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