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Shell Programming and Scripting

How to pass variables to FUNCTION ?

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# 1  
Old 05-31-2007
How to pass variables to FUNCTION ?
Hi...

Actually, I want to pass a few variables to the function and print it. But, its looks like not working. Could some body help me how could i do that ?... below is my program...


PHP Code:
#!/usr/bin/ksh

usage()
{
    echo "Usage: $0 -n -a -s -w -d"
    exit
}

rename()
{
    echo "rename $1 $2" 
}

if [ $# -lt 1 ]; then
    usage
fi

while getopts r:i:o opt
do
    case "$opt" in
        r) rename=1;;
        i) input="$OPTARG";;
        o) output="$OPTARG";;
        \?) usage;;
    esac
done

if [ $rename ]; then
    rename $input $output
fi 
# 2  
Old 05-31-2007
Try this...
This is what you want

myfunc()
{
echo "read var1 "$1
echo "read var1 "$2
}

myfunc 1 2
# 3  
Old 05-31-2007
As example:
lets say i run it as
#./script.sh -r -i input -o output

Then, the programm will passed "input" and "ouput" to rename function and print it as below:-
# rename input output

Now its blank and not print anything...

Anyway I have tried your solution but its same... and result has been print out..
# 4  
Old 05-31-2007
hmm... i have got it...

....Something wrong with the getopt declaration.....
# 5  
Old 05-31-2007
Your color coding is excellent.
# 6  
Old 05-31-2007
The colon must follows the option
Code:
while getopts ri:o: opt

Jean-Pierre.
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