String replace in unix

Question

I am to remove some text after a number for example: 1.2.0_testing. I just want the number part which is 1.2.0. SO anything after the underscore needs to be removed. Below is the code I am using:

echo $str | sed 's/_*//'

However the wildcard doesn't seem to work and I still get the full string. Can someone please help me.

Thanks!

That's not a wildcard, it's a regular expression!

ghoti
– ghoti

2016-05-05 11:19:56 +00:00
Commented May 5, 2016 at 11:19 — ghoti
– ghoti, Commented May 5, 2016 at 11:19

ohmu · Accepted Answer · 2016-05-05 15:09:07Z

4

Try this,

echo $str | sed 's/_.*//'

. - Matches any single character.

* - Matches zero or more occurrences of the previous character

So, _.* would represent 0 or more characters following _.

edited May 5, 2016 at 15:09

ohmu

19.9k44 gold badges114 silver badges151 bronze badges

answered May 5, 2016 at 11:13

Vadiraja K

8598 silver badges8 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Jens Over a year ago

Forking sed to remove a few characters from the end of a shell string is an anti-pattern that should not be proliferated.

Jens · Accepted Answer · 2016-05-05 11:21:59Z

3

No need for expensive forks to sed, awk, or - gasp! - perl. The shell can do this nicely:

$ str=1.2.0_testing
$ echo ${str%%_*}     # Remove longest part from the right matching _*
1.2.0

Learn all there is about built-in string manipulation with %, %%, #, ## in your fine shell manual. These remove from the right (percent) or left (hash) the shortest (single) or longest (double) match.

edited May 5, 2016 at 11:21

answered May 5, 2016 at 11:16

Jens

73.3k16 gold badges134 silver badges186 bronze badges

Collectives™ on Stack Overflow

String replace in unix

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related