How grep with -e (regex) `/log/messages` ? [ solved ]

Hi,

I would like to display the new lines of /var/log/messages that contain either IN_MyText or OUT_MyText (no matter where in the line)

I've tried

tail -fn 3 /var/log/messages | grep --color --line-buffered -e "(IN|OUT)_MyText"

But the output stay blank, when it should not...

Any ideas ?

5 comments

grep by default uses Basic Regular Expressions. This means the ( and ) lose their special meaning and are matched literally. Either use a backslash version \( to have a group, or use Extended Regular Expressions with -E "(IN|OUT)" . In man grep under REGULAR EXPRESSIONS are some differences noted.
- Thank you ! @thingsiplay@beehaw.org 👍
  -E solved it :)
It's marked solved, but since OP didn't post the solution:

-e uses basic regular expressions, where you need to escape the meta-characters ((|)) with a backslash. Alternatively, use extended regex with -E

$ echo a | grep -E "(a|b)" a $ echo a | grep -e "$a\|b$" a $ echo a | grep -e "(a|b)" $ echo a | grep -E "$a\|b$"
Do you get output if you use that exact tail command without the grep pipe?
- Yes

How grep with -e (regex) /log/messages ? [ solved ]