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# 1  
Old 11-01-2007
Help in getopts


My script will take a input file as a parameter(which is not mandatory) and also an option.
ksh <inputfile> [-n]
The option -n I have given is no way related to the input file.

Now the problem here is when i execute the script specifying the input file and the option(the way as mentioned above), the option part of the code is not working resulting me with no action.
But when i execute the script in this way, I am getting my desired output
ksh -n <inputfile>
But this is not the way i should mention them.

Why the option given at the end is not performing its action.? Is it a rule for getopts or problem with my code?

Please help me regarding this.

Sample code:

#! /bin/ksh
while getopts 'n' option
case "$option" in
n)echo "option is n"

?) echo " Bad option specified...."

shift `expr $OPTIND - 1`

Thanks in advance,
# 2  
Old 11-01-2007
For getopts the options must always be specified before the arguments.
The syntax for your script must be : ksh [-n] <inputfile>

# 3  
Old 11-01-2007
Thank you very much for the reply.

Can you explain me wat happens wen the option is given at the last?

ksh -n

In this case also I have given the option at the last. This works fine. Am I in the rite track?

where is the problem actually?

# 4  
Old 11-01-2007
If the option is specified after a normal argument, it is considered also as an argument.

In your case, you haven't specified the inputfile, the string "-n" is immediately after the command name so it is taken as an option.

# 5  
Old 11-02-2007
Originally Posted by aigles
If the option is specified after a normal argument, it is considered also as an argument.

In your case, you haven't specified the inputfile, the string "-n" is immediately after the command name so it is taken as an option.

Thank You for the reply.


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