bash IF not matching variable that contains regex numbers

Question

DPHPV = /usr/local/nginx/conf/php81-remi.conf;

I am unable to figure out how to match a string that contains any 2 digits:

if [[ "$DPHPV" =~ *"php[:digit:][:digit:]-remi.conf"* ]]

If you are trying to match files with names like php53-remi.conf then one way to do it is if [[ $DPHPV == php[[:digit:]][[:digit:]]-remi.conf ]]. — pjh
– pjh, Commented Jun 26, 2022 at 19:04
The whole purpose of quoting in Bash is to suppress the special significance that certain characters or character sequences would otherwise have, such as for globbing. — John Bollinger
– John Bollinger, Commented Jun 26, 2022 at 19:52
You're also mixing syntax for globbing, by using *, but =~ requires regular expressions, where the equivalent would be .*. Regexes alo aren't anchored, so the */.* can just be skipped. Glob expressions (==/= instead of =~) require matching the entire string, so there you would need the * — Benjamin W.
– Benjamin W., Commented Jun 26, 2022 at 19:54
The part of the rhs of the =~ , which is written between quotes, is interpreted as a literal match, not as a regular expression. Further, the lone * in front of the regex does not have any meaning as regex. You can see it when doing i.e. a [[ x =~ * ]], which does not match. — user1934428
– user1934428, Commented Jun 27, 2022 at 6:35

Wiktor Stribiżew · Accepted Answer · 2022-06-26 20:50:23Z

1

You are not using the right regex here as * is a quantifier in regex, not a placeholder for any text.

Actually, you do not need a regex, you may use a mere glob pattern like

if [[ "$DPHPV" == *php[[:digit:]][[:digit:]]-remi.conf ]]

Note

== - enables glob matching
*php[[:digit:]][[:digit:]]-remi.conf - matches any text with *, then matches php, then two digits (note that the POSIX character classes must be used inside bracket expressions), and then -rem.conf at the end of string. See the online demo:

#!/bin/bash
DPHPV='/usr/local/nginx/conf/php81-remi.conf'
if [[ "$DPHPV" == *php[[:digit:]][[:digit:]]-remi.conf ]]; then
    echo yes;
else
    echo no;
fi

Output: yes.

answered Jun 26, 2022 at 20:50

Wiktor Stribiżew

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