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up vote 1 down vote favorite

I would to grep certain parts of some shell command output in the shell script:

$ uname -r
>> 3.14.37-1-lts

Where I just need the 3.14.37.

And also for the shell script variable VERSION that has the value "-jwl35", I would like to take only the value "jwl35".

How can I use regular expression to this in shell script?

Thanks in advance!

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2 Answers 2

up vote 7 down vote accepted

Many, many ways. Here are a few:

  1. GNU Grep

    $ echo 3.14.37-1-lts | grep -oP '^[^-]*'
3.14.37

  2. sed

    $ echo 3.14.37-1-lts | sed 's/^\([^-]*\).*/\1/'
3.14.37

  3. Perl

    $ echo 3.14.37-1-lts | perl -lne '/^(.*?)-/ && print $1
3.14.37

    or

    $ echo 3.14.37-1-lts | perl -lpe 's/^(.*?)-.*/$1/'
3.14.37

    or

    $ echo 3.14.37-1-lts | perl -F- -lane 'print $F[0]'
3.14.37

  4. awk

    $ echo 3.14.37-1-lts | awk -F- '{print $1}'
3.14.37

  5. cut

    $ echo 3.14.37-1-lts | cut -d- -f1
3.14.37

  6. Shell, even!

    $ echo 3.14.37-1-lts | while IFS=- read a b; do echo "$a"; done
3.14.37
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Thank you soooo much! –  Jialun Liu Mar 29 at 15:55
    
what is the purpose of "-oP" in "grep -oP" ? –  Abdul Al Hazred Mar 29 at 16:03
1  
@AbdulAlHazred, "-o" is for printing only the matching part and "-P" is to use the Perl regex pattern. You can check for "man grep" for details –  Jialun Liu Mar 29 at 16:09
    
ok , got it, so without -o the whole line is shown , with -o only the substring that matches it. –  Abdul Al Hazred Mar 29 at 18:29
1  
@AbdulAlHazred exactly. It's a neat way to only print the relevant parts of a line. –  terdon♦ Mar 29 at 18:38
up vote 0 down vote

One approach not covered by terdon's comprehensive answer is Bash's parameter expansion (which requires no additional process):

vers=$(uname -r) && printf "%s\n" "${vers%%-*}"
3.14.37

vers="-jwl35" && printf "%s\n" "${vers#*-}"
jwl35
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