检查一行不是以grep开头的特定字符串 [英] Check if a line does not start with a specific string with grep

问题描述

我有一个文件app.log

  Oct 06 03:51:43 test test 
 Nov 06 15 ：04：53 text text 
 more text more text 
 Nov 06 15:06:43 text text text 
 Nov 06 15:07:33 
 more text more text 
 Nov 06 15:14:23 test test 
 more text more text 
 some more text 
 Nothing but text 
 some extra text 
 Nov 06 15:34： 31测试测试测试
  

如何将所有不在06年11月开始的行列入grep？

我试过了

  grep -En^ [^ Nov 06] app.log 
  

我无法获取其中包含06的行。

解决方案

只需使用下面的grep命令，

  grep -v'^ Nov 06'file 
  

从 grep - 帮助，

  -v，--invert-match选择不匹配的行
  

另一个通过正则表达式的破解，

  grep -P'^（?! Nov 06）'file 
  

正则表达式解释：

• ^


• （?! Nov 06）这个负面预测表明：在行开始之后没有字符串 Nov 06 。如果是，则匹配边界在每行的第一个字符之前存在。

另一个基于正则表达式的解决方案通过PCRE动词 （* SKIP）（* F）

  grep -P'^ Nov 06（* SKIP）（* F）| ^'file 
  

I have a file app.log

Oct 06 03:51:43 test test
Nov 06 15:04:53 text text text 
more text more text
Nov 06 15:06:43 text text text
Nov 06 15:07:33
more text more text
Nov 06 15:14:23  test test
more text more text
some more text 
Nothing but text
some extra text
Nov 06 15:34:31 test test test

How do I grep all the lines that does not begin with Nov 06 ?

I have tried

grep -En "^[^Nov 06]" app.log

I am not able to get lines which have 06 in them.

解决方案

Simply use the below grep command,

grep -v '^Nov 06' file

From grep --help,

-v, --invert-match        select non-matching lines

Another hack through regex,

grep -P '^(?!Nov 06)' file

Regex Explanation:

• ^ Asserts that we are at the start.
• (?!Nov 06) This negative lookahead asserts that there isn't a string Nov 06 following the line start. If yes, then match the boundary exists before first character in each line.

Another regex based solution through PCRE verb (*SKIP)(*F)

grep -P '^Nov 06(*SKIP)(*F)|^' file

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