Bash scripting, use regex as a case pattern

[paolodina]

Hi, referring to this syntax of case:
case word in
[(]pattern1) compound-list;;
esac

when pattern1 is a regular expression with expressions grouped using parentheses like in:
(25[0-5]|2[0-4][0-9]|[0-1]?[0-9]?[0-9])(\.(25[0-5]|2[0-4][0-9]|[0-1]?[0-9]?[0-9]))\{3\}

an error occours, probably due to the ')' after the regular expression.
-bash: ./badscript: line 120: syntax error near unexpected token `25[0-5]'
-bash: ./badscript: line 120: ` (25[0-5]|2[0-4][0-9]|[0-1]?[0-9]?[0-9])(\.(25[0-5]|2[0-4][0-9]|[0-1]?[0-9]?[0-9]))\{3\} )'

How fix it?

Thanks

[drl]

Hi, paolodina.

I have no immediate solution for you without some thought, but the expressions used in the case statement are essentially pathname expressions (such as you use on the command line), not regular expressions:

       case word in [ [(] pattern [ | pattern ] ... ) list ;; ] ... esac
              A case command first expands word, and tries to match it against
              each pattern in turn, using the same matching rules as for path-
              name  expansion (see Pathname Expansion below).  When a match is
              found, the corresponding list  is  executed.   After  the  first
              match,  no subsequent matches are attempted.  The exit status is
              zero if no pattern matches.  Otherwise, it is the exit status of
              the last command executed in list -- man bash

You may need to do some processing with sed if you want to involve regular expressions in a case statement.

Perhaps our colleagues have some more creative ideas ... cheers, drl

[skog]

case $name in
[ab]{2,3} )
echo do stuff
;;
* )
;;
esac