Bash regexp in if statement doesn't work [duplicate]

Question

I'm trying to play with regular expressions in bash but i can't understand why this follow scenario doesn't work :

Regexp:

REGEXP="^(test\/)(([a-zA-Z]+)\-)+(0|[1-9]\d*)\.(0|[1-9]\d*)\.(0|[1-9]\d*)$"

String:

STRING="test/test-ods-1.10.1"

Test:

if [[ "$STRING" =~ $REGEXP ]]
then
       echo "match!"
else
       echo "don't match"
fi

Normally in this scenario I should receive a "match" but it's always returning a "doesn't match".

Just a small hint [[ "foo" =~ foo ]]; echo $? shows 0 in case of a match, 1 if not, and 2 if there is a syntax error in the regex. — Micha Wiedenmann
– Micha Wiedenmann, Commented Sep 2, 2016 at 9:36

Aaron · Accepted Answer · 2016-09-02 09:48:28Z

3

\d is not defined in bash regexps. Replace them with [0-9] and it'll work :

$ [[ "test/test-1.10.1" =~ ^(test\/)(([a-zA-Z]+)\-)+(0|[1-9]\d*)\.(0|[1-9]\d*)\.(0|[1-9]\d*)$ ]]; echo $?
1

$ [[ "test/test-1.10.1" =~ ^(test\/)(([a-zA-Z]+)\-)+(0|[1-9][0-9]*)\.(0|[1-9][0-9]*)\.(0|[1-9][0-9]*)$ ]]; echo $?
0

No shorthand classes are defined in POSIX, and GNU extensions only bring a few of them, \w, \W, \s and \S according to regular-expressions.info.

edited Sep 2, 2016 at 9:48

answered Sep 2, 2016 at 9:42

Aaron

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