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How to pass variables to FUNCTION ? How to pass variables to FUNCTION ?
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  #1  
Old 05-30-2007
Registered User
  

Join Date: May 2007
Posts: 6
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 
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  #2  
Old 05-30-2007
Registered User
  

Join Date: May 2007
Posts: 13
Try this...
This is what you want

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

myfunc 1 2
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  #3  
Old 05-30-2007
Registered User
  

Join Date: May 2007
Posts: 6
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..
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  #4  
Old 05-30-2007
Registered User
  

Join Date: May 2007
Posts: 6
hmm... i have got it...

....Something wrong with the getopt declaration.....
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  #5  
Old 05-30-2007
Registered User
  

Join Date: Jan 2005
Posts: 212
Your color coding is excellent.
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  #6  
Old 05-30-2007
aigles's Avatar
Registered User
  

Join Date: Apr 2004
Location: Bordeaux, France
Posts: 1,212
The colon must follows the option
Code:
while getopts ri:o: opt
Jean-Pierre.
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