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Problem in bash script

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# 1  
Old 11-18-2010
Problem in bash script
I have written a script and I get error and I don't understand why.
Code:
neededParameters=2
numOfParameters=0
correctNum=0
while getopts "s:l:" opt 
do
    case "$opt" in
    s)
        serviceName= $OPTARG #errorline 1 
        numOfParameters= $numOfParameters + 1
        ;;
    l)
        fileName= $OPTARG #errorline 2 
        numOfParameters= $numOfParameters + 1    
        ;;
    ?)
        echo "unknown option"
        ;;
    esac
done

when I run:
script -s s -l l
Then I get error message on errorlines 1 and 2 that it doesn't recognize s and l.

I would be really glad for an help.
Thanks in advance.
# 2  
Old 11-18-2010
ok i tested this under linux, it runs fine

Code:
neededParameters=2
numOfParameters=0
correctNum=0
while getopts "s:l:" opt
do
    case $opt in
    s)
        serviceName=$OPTARG #errorline 1
        let numOfParameters++
        ;;
    l)
        fileName=$OPTARG #errorline 1
        let numOfParameters++
        ;;
    \?)
        echo "unknown option"
        ;;
    esac
done

Last edited by ctsgnb; 11-18-2010 at 11:31 AM.. Reason: need testing before
This User Gave Thanks to ctsgnb For This Post:
programAngel
# 3  
Old 11-18-2010
Modified version:

Code:
neededParameters=2
numOfParameters=0
correctNum=0
while getopts "s:l:" opt
do
    case $opt in         # the quotes are unnecessary in this context
    s)
        serviceName=$OPTARG # no space after the equal sign
        numOfParameters=$(( $numOfParameters + 1 )) # I suppose you need arithmetic expansion here
        ;;
    l)
        fileName=$OPTARG #errorline 2 
        numOfParameters=$(( $numOfParameters + 1 )) 
        ;;
    ?)
        echo "unknown option"
        ;;
    esac
done

This User Gave Thanks to radoulov For This Post:
programAngel
# 4  
Old 11-18-2010
Sorry I could not use s (for there is a silly program with that name on this box (AIX)...)
Code:
an12:/home/vbe $ more bash_opt.sh
neededParameters=2
numOfParameters=0
correctNum=0
#while getopts "s:l:" opt 
while getopts t:l: opt 
do
    case "$opt" in
    t)
        serviceName=$OPTARG #errorline 1 
        numOfParameters=`expr $numOfParameters + 1`
        ;;
    l)
        fileName= $OPTARG #errorline 2 
        numOfParameters=`expr $numOfParameters + 1`
        ;;
    ?)
        echo "unknown option"
        ;;
    esac
done
echo neededParameters= $neededParameters
echo numOfParameters= $numOfParameters
an12:/home/vbe $ bash -debug bash_opt.sh -t t -l l
bash_opt.sh: line 13: l: command not found
neededParameters= 2
numOfParameters= 2

This User Gave Thanks to vbe For This Post:
programAngel
# 5  
Old 11-18-2010
Quote:
Originally Posted by ctsgnb
should be :s:l
instead of s:l:
?
O.K.

But why do you think that?

Because from what I understand when there is colons after the letter then it mean that a parameter must come after it when you run the script.
I mean it mean that if you write in the shell:
script -s
then the colons after the s (in the script) will send error.
Hope I explain myself good.
# 6  
Old 11-18-2010
Quote:
Originally Posted by ctsgnb
should be :s:l
instead of s:l:
?
These two statements have different meanings.
# 7  
Old 11-18-2010
Quote:
Originally Posted by vbe
Sorry I could not use s (for there is a silly program with that name on this box (AIX)...)
Code:
an12:/home/vbe $ more bash_opt.sh
neededParameters=2
numOfParameters=0
correctNum=0
#while getopts "s:l:" opt 
while getopts t:l: opt 
do
    case "$opt" in
    t)
        serviceName=$OPTARG #errorline 1 
        numOfParameters=`expr $numOfParameters + 1`
        ;;
    l)
        fileName= $OPTARG #errorline 2 
        numOfParameters=`expr $numOfParameters + 1`
        ;;
    ?)
        echo "unknown option"
        ;;
    esac
done
echo neededParameters= $neededParameters
echo numOfParameters= $numOfParameters
an12:/home/vbe $ bash -debug bash_opt.sh -t t -l l
bash_opt.sh: line 13: l: command not found
neededParameters= 2
numOfParameters= 2

Thanks I will check your suggestion.

---------- Post updated at 04:53 PM ---------- Previous update was at 04:46 PM ----------

Code:
numOfParameters=`expr $numOfParameters + 1`

Is this the way to do arithmetic in bash script.
using Quotation markand and exper.
Because I thought you can just write the variables and do use math operator.
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