bash regex in if statement

Question

I'm trying to add name-checking with regular expression, that passes only characters and digits but no special symbols given from the output. I wrote this code, but it's does not working. It shows "not ok" either when I'm typing only characters with digits or chars+special symbols

#!/bin/bash
regex="/^[a-zA-Z0-9_]+$/gm"
read -p "Type smth: " text

if [[ $text =~ $regex ]]
then
    echo "ok"
else
    echo "not ok"
fi

Here's the output:

user@localhost:~/Documents/scripts$ ./testregex.sh
Type smth: hello$#!
not ok

user@localhost:~/Documents/scripts$ ./testregex.sh
Type smth: hello
not ok

Remove surrounding slashes and the gm option from your regex as regex="^[a-zA-Z0-9_]+$". — tshiono
– tshiono, Commented Jul 6, 2022 at 10:59
Your regex is actually Perl-style matching operator, not just a regular expression. — chepner
– chepner, Commented Jul 6, 2022 at 22:15

Wiktor Stribiżew · Accepted Answer · 2022-07-06 10:16:27Z

2

You can use

if [[ $text =~ ^[[:alnum:]_]+$ ]]
then
    echo "ok"
else
    echo "not ok"
fi

Details:

^ - start of string
[[:alnum:]_]+ - one or more letters, digits or underscores
$ - end of string.

Note the absence of regex delimiter chars.

See the online demo.

answered Jul 6, 2022 at 10:16

Wiktor Stribiżew

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