Why is this simple bash script not working? getopts

Question

I don't get why this won't work as it's really simple.

#!/bin/bash
type='A'

while getopts "t:" OPTION
do
  case $OPTION in
    t)
       echo "The value of -t is $OPTARG"
       type=$OPTARG
       exit
       ;;
    \?)
       echo "Used for the help menu"
       exit
        ;;
  esac
done
echo $type

Output I get:

root@w:/etc/scripts# ./dns_add_record -t test  
The value of -t is test  
root@w:/etc/scripts# ./dns_add_record  
A

Expected output:

root@w:/etc/scripts# ./dns_add_record -t test  
The value of -t is test  
test

Can someone figure out what is wrong? It's probably something stupid, but I can't get this working the way I want it to.

Karoly Horvath · Accepted Answer · 2016-01-02T10:38:42.480

2

exit exits from the shell script.

Remove it from the -t case, it only makes sense for help.

Adding set -x to your script gives this trace:

+ type=A
+ getopts t: OPTION
+ case $OPTION in
+ echo 'The value of -t is 3'
The value of -t is 3
+ type=3
+ exit

edited Jan 02 '16 at 10:38

answered Jan 02 '16 at 10:37

Karoly Horvath

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Wow... Used to java, so figured you needed it at the end of a case. Thanks! – woutwoot Jan 02 '16 at 10:38
I don't know why you've thought that, there's nothing similar in Java. – Karoly Horvath Jan 02 '16 at 10:44
1

@KarolyHorvath I think OP is referring to [fall through](http://stackoverflow.com/questions/5562253/switch-case-with-fallthrough). – user000001 Jan 02 '16 at 11:04

Why is this simple bash script not working? getopts

1 Answers1