使用 grep 检查一行是否不以特定字符串开头 [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

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

如何 grep 所有不是以 Nov 06 开头的行?

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

我试过了

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

我无法获取包含 06 的行.

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

推荐答案

只需使用下面的 grep 命令，

Simply use the below grep command,

grep -v '^Nov 06' file

来自 grep --help,

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

通过正则表达式的另一个黑客，

Another hack through regex,

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

正则表达式解释:

• ^ 断言我们处于开始阶段.
• (?!Nov 06) 这个否定前瞻断言在行开始之后没有字符串 Nov 06.如果是，则匹配每行中第一个字符之前存在的边界.

• ^ 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.

通过 PCRE 动词的另一种基于正则表达式的解决方案 (*SKIP)(*F)

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

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

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