ksh script


 
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# 1  
Old 07-14-2009
ksh script

I have a script like this ..

Code:
#!/usr/bin/ksh

if [ -z "${OUT_LOG_FILE_NAME}" ]
then  
  OUT_LOG_FILE_NAME=0
fi

LOGFILE="sample.log"

if [ ${OUT_LOG_FILE_NAME} -ne 0 ] 
then
	echo ${LOGFILE}
fi

what I am trying to do is if OUT_LOG_FILE_NAME is passed then it should return value of logfile name . It's returning value in all cases . what I am doing wrong here ?

Last edited by vgersh99; 07-14-2009 at 02:03 PM.. Reason: code tags, PLEASE!
# 2  
Old 07-14-2009
To keep the forums high quality for all users, please take the time to format your posts correctly.

First of all, use Code Tags when you post any code or data samples so others can easily read your code. You can easily do this by highlighting your code and then clicking on the # in the editing menu. (You can also type code tags [code] and [/code] by hand.)

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Thank You.

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# 3  
Old 07-14-2009
By "if OUT_LOG_FILE_NAME is passed" do you mean passed as an argument on the command line?
# 4  
Old 07-14-2009
Beware if OUT_LOG_FILE_NAME is set to a non-numeric value the "if -ne" will fail. Not clear whether you are using this variable as a boolean.
# 5  
Old 07-14-2009
hello,
I assume you are passing in a filename to the script, and if nothing is passed in you want to use the default filename of "sample.log".

Try this 2-liner:

LOGFILE=${1:-"sample.log"}
echo ${LOGFILE}

#This means if nothing was passed, use the default "sample.log".
# 6  
Old 07-14-2009
what I am trying to do is ..

$ ./1.sh OUT_LOG_FILE_NAME=0
0

./1.sh OUT_LOG_FILE_NAME=1
0


if I execute the script 1.sh with out OUT_LOG_FILE_NAME parameter , it should not return log file name from the program .

If I pass OUT_LOG_FILE_NAME , it should return log file name from 1.sh program .

Thanks
# 7  
Old 07-15-2009
example, testing argument 1
Code:
case "$1" in
     OUT_LOG_FILE_NAME=0) ;;
     OUT_LOG_FILE_NAME=1) echo "$LOGFILE" ;;
     *) ;;
esac

commandline
1.sh OUT_LOG_FILE_NAME=0
means that argument 1 value is "OUT_LOG_FILE_NAME=0", commandline not set value for variable.

If you like to use variable, then first set variable and then call script. Variable must be global, if you like to use it in childprocess. Using export you make global variable.
export OUT_LOG_FILE_NAME
OUT_LOG_FILE_NAME=0
./1.sh


If you like to give command line like *nix system has done in history, then something like
./1.sh -o 1
or
./1.sh --outlog 1
are "standard" methods.

How to do it ?
You can use command getopt, but here is example how to take full control of options and arguments:

Code:
# init flags/variables/...
flag1=0
# handle command line options and arguments
while [ $# -gt 0 ] 
do 
    arg="$1" 
    case "$arg" in 
          -o|--outlog) flag1="$2"
               shift ;; 
          --) # no more option
               shift
               break
               ;;
          -*) echo "usage..." >&2 
                exit 1 
                ;; 
          *) break ;; # no options anymore, this is already argument
    esac 
    shift 
done 

[ "$flag1" = "1" ] && echo "flag1 is 1"

# now you can handle all arguments
for arg in $*
do
    cmd "$arg"
done


Last edited by kshji; 07-15-2009 at 04:39 AM..
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