如果没有提供选项，运行子程序的最佳方法是什么？ Perl getOpt

Question

For example,

If I run

./program -i 

do subA

if I run

./program 

do subB

my code currently looks like this

subB if(!$opt_a && !$opt_b && !opt_b);

but this looks messy. Is there anyway to make it run subB if no options provided rather than check each individual option?

Answer 1

You can store the options in a hash and check to see if the hash evaluates to true or false:

use strict;
use warnings;

use Getopt::Long;

my %options;
GetOptions( \%options, 'opt_a=i', 'opt_b', 'opt_c' );

if ( %options ) {

    # Season with option validation ..
    # Usage : $options{opt_a} instead of $opt_a

    subA();
}

else {

    subB();
}

Answer 2

I sometimes use a delete hash idiom when a tool only operates on one (or specific combinations) of options so that I can let the user know if everything was consumed as expected. Update: just replace the print statements with your sub calls.

use strictures;
use Getopt::Long;

GetOptions( \my %opt, "int=i", "str=s", "bool" );

print "OPTIONS, I HAZ DEM\n" if %opt;

if ( delete $opt{int} )
{
    print "* I HAZ A INT\n";
}
elsif ( delete $opt{str} )
{
    print "* I HAZ STRING\n";
}
elsif ( delete $opt{bool} )
{
    print "* I HAZ A TRUTH\n";
}
else
{
    print "I CAN HAZ OPTION?\n";
}

print "DO NOT WANT: ", join(", ", keys %opt), $/
    if %opt;

Answer 3

I'd check each individual option anyway. It makes things much clearer for whoever has to modify the program in three years time... and it might even be you.

Bite the bullet and code it in full. Its tedious but saves time in the long run.

    if ($opt_a) 
  {
    # do whatever
  }
elsif ($opt_b)      
  {
    # do whatever
  }
else
  {
    # default behaviour if no arguments passed from command line
  }