You're probably looking for something like that:

my $dirtype = $in{username} =~ /^([a-zA-Z]{2})$/ ? $1 : 'other' ;

if ($in{username} =~ /^(\[a-zA-Z]{2})/ || ' ') {
     $dirtype = $1 || 'other';
}
[download]

Second of all, IF the regular expression DOES match (triggering the code between the brackets and populating $1 at the same time) THEN in your case $1 is always guaranteed to be true (because it will contain two A-Za-z letters. Letters always evaluate to "true." Because the first expression in your short-circuit operation ($1) is always true, your program will never get to the 'other' value.

This is why OeufMayo resorted to a ternary operator, which is just a short-hand way to write another if/then statement. Here it is rewritten for clarity:

my $dirtype;
if ( $in{username} =~ /^([a-zA-Z]{2})$/ ) {
     $dirtype = $1;
} else {
      $dirtype = 'other';
}
[download]

And then there's the backslash in your regex. :-D

Gary Blackburn
Trained Killer

You were right the second time. That is a good way to explain things. Usually it is best to start out with a multi-line piece of code, rather than trying to write a very complex one liner. It is easier to debug the multi-line piece and then simplify it down into a one liner.

One clarification... (Unfortunately, " "is NEVER true in Perl)... There are some things that are true that you wouldn't think are true (in this case " " actually is). Try some of the following...


perl -e 'print "1\n" if "";'
perl -e 'print "1\n" if " ";'
perl -e '@a=(); print "1\n" if @a;'
perl -e '@a=(undef); print "1\n" if @a;'
[download]


Empty strings and empty lists do evaluate to false. Strings and lists with even one item evaluate to true, even if that one item is itself false.

my @a=qw(random brilliant braindead); print $a[rand(@a)];

Thank you so much for the excellent logic lesson. I stripped it apart like you suggested, and now it is working wonderfully.

Thanks again for all the responses! You Perl Monks are such enlightened brethren! ; )

Sad Gurunath Maharaj ki jai!