Help on command line argument in csh

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09-27-2011

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Quote:

Originally Posted by scottn

Are you not missing the "case" keyword?

Code:

#!/usr/bin/csh
while ( $#argv != 0 )
   switch ( $argv[1] )
     case "-kit":
       KIT=$2;
       echo "Proc arg missing"
       shift
       breaksw
     case "-type":
       TYPE=$1;
       echo "TYPE is $1"
       breaksw
     case "-tl":
       TL=$1;
       echo "TL is $1"
       breaksw
     case "-PILX":
       PILX=$1;
       echo "PIL is $1"
       breaksw
     default:
       echo "Invalid arg"
       breaksw
   endsw
   shift
end

And why does it have to be C-Shell? Why not use a proper shell?!

Why do some options in the example command you posted not appear in the switch statement?

Sory for missing cases. Its not necessary to have all the options in a command.

vineet.dhingra

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