grep - 使用grep进行负匹配（匹配不包含foo的行）(Negative matching using grep (match lines that do not contain foo))

Question

I have been trying to work out the syntax for this command:

(我一直在试图找出这个命令的语法：)

grep ! error_log | find /home/foo/public_html/ -mmin -60

or

(要么)

grep '[^error_log]' | find /home/baumerf/public_html/ -mmin -60

I need to see all files that have been modified except for those named error_log .

(我需要查看已修改的所有文件，但名为error_log文件除外。)

I've read about it here , but only found one not -regex pattern.

(我在这里读过它 ，但只找到一个not -regex模式。)

  ask by jerrygarciuh translate from so

Answer 1

grep -v is your friend:

(grep -v是你的朋友：)

grep --help | grep invert  

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

(-v， - inverse-match选择不匹配的行)

Also check out the related -L (the complement of -l ).

(另请查看相关的-L （ -l的补码）。)

-L, --files-without-match only print FILE names containing no match

(-L， - files-without-match仅打印不包含匹配项的FILE名称)