Variable in firewall-cmd command

Question

I'm trying to do a script in Bash to add some firewall rules.

The variable I'm passing isn't working, I guess I'm doing the substitution wrong.

firewall-cmd --add-rich-rule='rule family="ipv4" source address="$IP/32" port port=10000 protocol=tcp accept'

Returns

Error: INVALID_ADDR: $IP/32

What's wrong here, and how do I fix it?

Please post the output of `echo $IP` to check the bash variable IP is correctly set. — Pierre François, Jan 12 '20 at 09:03
See: [Difference between single and double quotes in bash](http://stackoverflow.com/q/6697753/3776858) — Cyrus, Jan 12 '20 at 10:14

score 0 · Accepted Answer · edited Apr 13 '21 at 05:38

0

You are single-quoting the entire string, so you are passing the literal string $IP. You can break out of the single quotes temporarily by adding a closing single quote before the variable and a new single quote after it.

firewall-cmd --add-rich-rule='rule family="ipv4" source address="'$IP'/32" port port=10000 protocol=tcp accept'

edited Apr 13 '21 at 05:38

tripleee

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answered Jan 12 '20 at 09:09

Nitish

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Variable in firewall-cmd command

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