[SOLVED] bash : getopts problem in bash script.

angel115 · 03-02-2009, 09:53 AM

Hello There,

I'm having an issue using getopts in my bash script.
For trouble shooting purposes I've make it simple:

Code:

#!/bin/bash

while getopts ":f:p:d:g:a" OptionArgument; do
case $OptionArgument in
	
	p ) productdb=${OPTARG};;
	d ) debug="1";;
	f ) cdtFile=${OPTARG};;
	g ) generic="1";;
	a ) a="just another one just for test";;
	\?) echo = "No option specify";;
	* ) echo = "wrong variable";;
esac
done


echo "product = ${productdb}"
echo "cdtFile = ${cdtFile}"
echo "debug = ${debug}"
echo "generic = ${generic}"
echo "a= ${a}"

In my example p and f will be optional and a is just for testing.

below is an example of what I type and the output:

Code:

$ ./1_analysing -d -g -a -pimsx -ffileName 
product = imsx
cdtFile = fileName
debug = 1
generic = 
a= just another one just for test

the generic value stay empty

which I don't why.

I've found many example on internet but none of them are working as expected.
Is there a bug some where or did I miss something?

Thanks for your help;
Angel.

bgeddy · 03-02-2009, 10:46 AM

Try it with this change :

Code:

while getopts ":f:p:dga" OptionArgument; do

From getopts help :

Quote:

if a letter is followed by a colon, the option is expected to have an argument, which should be separated from it by white space.

angel115 · 03-02-2009, 10:53 AM

Quote:

Originally Posted by bgeddy

Try it with this change :

Code:

while getopts ":f:p:dga" OptionArgument; do

From getopts help :

Hehe, you save my life thank you so much, my script is working like a charm now.

We can even go simple: (which after testing work fine as well)

Code:

while getopts "f:p:dga" OptionArgument; do

Angel.