Regex in Bash "if" Statement

Question

if [[ $1  =~ ^\'*\"*[\d+\w+]+\'*\"*$ ]]
    then
        echo true
    else
        echo error
        exit 1
fi

This regex seems to be correct (verified with https://regexr.com, and other regex sites), however, the script will evaluate to false. Any idea why?

For this argument, I am expecting a line such as:

Will0w

To be matched, however, no luck.

Any help would be appreciated,

Thanks

"verified with regexr.com, and other regex sites": here is your error, bash regexes don't use the same syntax (and don't have the same behaviour) than javascript or php patterns. Search about the ERE (Extended Regular Expression) syntax. — Casimir et Hippolyte
– Casimir et Hippolyte, Commented Mar 4, 2018 at 19:50

Gilles Quénot · Accepted Answer · 2018-03-04 22:51:41Z

2

Using bash :

if [[ $1  =~ ^\'*\"*[[:alnum:]]+\'*\"*$ ]]
    then
        echo true
    else
        echo error
        exit 1
fi

edited Mar 4, 2018 at 22:51

answered Mar 4, 2018 at 20:16

Gilles Quénot

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1 Comment

chepner Over a year ago

[a-Z] depends on [a-z] and [A-Z] being adjacent in the current locale. Use the POSIX character class [[:alnum:]] instead, which also includes 0-9.

albetancourt · Accepted Answer · 2018-03-04 20:06:33Z

1

An alternative solution using grep:

if echo $1 | grep -qP "^\'*\"*[\d+\w+]+\'*\"*$"
    then
        echo true
    else
        echo error
        exit 1
fi

answered Mar 4, 2018 at 20:06

albetancourt

362 bronze badges

1 Comment

codeforester Over a year ago

grep -qP "^\'*\"*[\d+\w+]+\'*\"*$" <<< "$1" will be more efficient as it eliminates the fork needed for the pipe.

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