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Linux is a family of open source Unix-like operating systems based on the Linux kernel, an operating system kernel first released on September 17, 1991 by Linus Torvalds. Linux is typically packaged in a Linux distribution (or distro for short).

Distributions include the Linux kernel and supporting system software and libraries, many of which are provided by the GNU Project. Many Linux distributions use the word "Linux" in their name, but the Free Software Foundation uses the name GNU/Linux to emphasize the importance of GNU software, causing some controversy.

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

submitted 1 month ago* (last edited 1 month ago) by Gordon_F@lemmy.ml to c/linux@lemmy.ml

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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 ?

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[–] CosmicGiraffe@lemmy.world 4 points 1 month ago

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 "\(a\|b\)"