Hi guys I have a Unix Shell script that is a wrapper to a stored procedure on an Oracle database. The script is invoked by Maestro and is passed a parameter. The parameter is a date in the format DD/MON/YYYY (uk format), and I was wondering how I should validate the format within the script. I know how to test for a single argument being passed in:

if [ $# -ne 1 ]; then

  echo 'USAGE: script_name <effective_frm_date>'
  echo         effective_frm_date format - DD/MON/YYYY'
  exit 1
fi;

But I have no idea how to test the actual format of teh parameter. Has anyone any ideas? Thanks

Look into sed and awk. Both are tools for pattern matching. It's a good idea to learn to use them anyway in case you need to write more non-trivial shell scripts in the future.

I've discovered awk...very handy.

A quick question though...if I have the following in a shell script:

echo $DATE $LOGFILE | awk '{               if ((substr($1,3,1) != "/") || (substr($1,6,1) != "/"))               {                    print "USAGE: brs1_valid_data.ksh <efftve_from_date>" > $2                    print "       efftve_from_date format - DD/MM/YYYY" >> $2                    print "       Invalid Date Format" >> $2                    exit 1;               }              }'DAY=`echo $DATE | awk '{FS="/"} {print $1}'`MONTH=`echo $DATE | awk '{FS="/"} {print $2}'`YEAR=`echo $DATE | awk '{FS="/"} {print $3}'`echo $DAY $MONTH $YEAR $TODAY_DAY $TODAY_MONTH $TODAY_YEAR $LOGFILE | awk '{                if(($3 > $6) ||                   (($3 == $6) && ($2 > $5)) ||                   (($3 == $6) && ($2 == $5) && ($1 > $4)))                {                    print "USAGE: brs1_valid_data.ksh <efftve_from_date>" > $7                    print "       efftve_from_date format - DD/MM/YYYY" >> $7                    print "       Date cannot be in future" >> $7                    exit 1;                }               }'

Will the exit(s) within the awk commands exit from the script OR simply from the awk command?

EDIT: Okay have realised that I need to check the $? value after the awk commands.

[Edited by - garyfletcher on January 19, 2006 10:11:26 AM]