How can remove suffix including hyphen and number in Linux filename?

Question

I have some text file like this

access-2018-08-21.log.1-20180906
access-2018-08-22.log.1-20180906

I want to remove the -20180906 part wo the result would be

access-2018-08-21.log
access-2018-08-22.log

I tried rename -- "s/\-20180906//g" * but it didn't work.

What's the command line i need to achieve my goal?

Wrong rename utility? See Why is the rename utility on Debian/Ubuntu different than the one on other distributions, like CentOS? and check what rename --version says. — ilkkachu, Sep 06 '18 at 08:26
Does not work in what way? Also, your command is a bit wrong. Something like the following suffices: rename "s/\.1-20180906//g" * — Sparhawk, Sep 06 '18 at 08:26
Hi, it didn't work in a way that doesn't remove the -20180906 part. And by the way, your above-mentioned command also doesn't remove the -20180906 part. — The One, Sep 06 '18 at 08:29
Sorry, I re-edited the comment. But what is the precise output when it doesn't work? — Sparhawk, Sep 06 '18 at 08:31
@Sparhawk It didn't have any output. After executing your command, i used ls to check and the file name still remained. — The One, Sep 06 '18 at 08:33
Thanks. And as per the other comment, what is the output of rename --version? — Sparhawk, Sep 06 '18 at 08:33
Okay, thanks. So @ikkachu's link is spot on. Have a read of that. You need the perl version of rename instead. — Sparhawk, Sep 06 '18 at 08:35

Kamaraj · Answer 1 · 2018-09-06T09:54:52.780

1

Try this..

if you are happy with the output, then just remove the echo word.

for i in access*; do echo mv $i ${i%.*}; done

edited Sep 06 '18 at 09:54

answered Sep 06 '18 at 08:43

Kamaraj

Just started using the {...} expansions in some of my scripts. It's much faster than spawning a new process with $(...) – user208145 Sep 06 '18 at 09:27
1

@Kamaraj s/cp/mv/ – Siva Sep 06 '18 at 09:42

1 Answers1