my @sorted = sort {
    $hash{$a}->{employee_number}->{years_employed}
    <=>
    $hash{$b}->{employee_number}->{years_employed}
} keys %hash;
[download]

_________
broquaint

Is there any way to do this without creating a new array of the sorted values? I really do appreciate the help, but that wasn't what I was hoping for.

Hashes themselves cannot be sorted -- the closest one can get is to sort either the keys or the values of the hash, then iterate through them. See perldoc -q "sort a hash"

perl -pe '"I lo*`+$^X$\"$]!$/"=~m%(.*)%s;$_=$1;y^`+*^e v^#$&V"+@( NO CARRIER'

Take broquaint's answer, remove the assignment, add a loop and a print. Job done.

Or do you just want the code:

my %hash;
while (<DATA>) {
   my @data = split;
   $hash{$data[0]}->{employee_number} = $data[1];
   $hash{$data[0]}->{years_employed} = $data[2];
}
foreach (sort {
              $hash{$a}->{years_employed}
              <=>
              $hash{$b}->{years_employed}
              } keys %hash) {
   print "$_ $hash{$_}->{employee_number} $hash{$_}->{years_employed}\
+n";
};

__DATA__
SS123 EMP123 3
SS234 EMP124 2
SS456 EMP125 1
[download]

prints

SS456 EMP125 1
SS234 EMP124 2
SS123 EMP123 3
[download]

rdfield

foreach my $ss_number {keys %my_hash} {
  print "$my_hash{$ss_number}{$employee_number}\n";
}
[download]

Your data structure doesn't look right to me, unless you really want to have multiple employee_numbers per ss_number.

Probably you want a hash with ss_number as the key and the value as either an array ref or hash ref with all the values associated with that ss_number. E.g.

 $myhash{'555236801'} = [3035, 2];
or
 $myhash{'555236801'} = { empno => 3035, years => 2 };
[download]

Anyway, one easy way to get what you want is like this (at least on Unix (using the latter data structure:

  open SORTEDOUT, "|sort +2" or die "cannot open sort: $!\n";
  for (keys %myhash) {
    print SORTEDOUT $_, " ", $myhash{$_}{empno}, " ", $myhash{$_}{year
+s}, "\n";
  }
  close (SORTEDOUT) or die "Error on sort: $!\n";
[download]