in reply to Validating command-line _parameters_ (not options)

Your test will look for a "false" value of either thing. That use of && itself triggers the warning! Say 
if (!defined($username) || !defined($dominio))

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for the effect you want.

Or, I'll just report on how many arguments there are in the first place. 

if (scalar(@ARGV) != 2)

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(more often a range of legal values)

—John

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Re: Re: Validating command-line _parameters_ (not options)
by dserodio (Novice) on Sep 15, 2001 at 02:10 UTC
    I tried using
    if (!defined ($username)
    || !defined ($domain))

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    But I'm still getting the same warnings.
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[d/l]
      Maybe the split is causing the warning if ARGV[0] is undefined. Check for the presence of a parameter first!

      make sure @ARGV contains exactly one element.
      split ARGV[0] at the @
      Make sure the split gave you both parts.

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[d/l]
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