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bash : getopts problem in bash script.

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.

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.

Quote:

Originally Posted by bgeddy (Post 3462513)

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. :cool:

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

Code:

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

Angel.