[SOLVED] bash scripting options double semicolon

secomax · 02-26-2017, 07:12 AM

Hi
I have a small question about bash scripting
this is the snippet I'm asking about

Code:

#!/bin/bash
echo
while [ -n "$1" ]
do
case "$1" in
-a) echo "Found the -a option" ;;
-b) echo "Found the -b option" ;;
-c) echo "Found the -c option" ;;
*) echo "$1 is not an option" ;;
esac
shift
done

I got this from a tutorial https://likegeeks.com/linux-bash-scr...e-guide-part3/

Why do we need double semicolon after each option
I tried to remove the semicolon it gives me this error

Code:

syntax error near unexpected token `)'

thanks in advance.

hydrurga · 02-26-2017, 07:33 AM

You need the double semicolon at the end of each case clause because that is what bash syntax requires in order to parse the command correctly.

http://tldp.org/LDP/Bash-Beginners-G...ect_07_03.html

Remember that each case clause may have multiple statements within that clause, each terminated with an optional semicolon. Bash needs to know when the clause itself is finished, hence the double semicolon.

secomax · 02-26-2017, 08:37 AM

thanks for the reply.

hydrurga · 02-26-2017, 08:48 AM

Glad to help, secomax.

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