Something like this (assuming 'count' contains a numeric value):

my( @sorted ) = sort { 
    $users{$a}->{count} <=> $users{$b}->{count}
} keys %users;
print "$_ => $users{$_}->{count}\n" foreach @sorted;
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And if it isn't numeric, replace the '<=>' (spaceship operator) with 'cmp' .

...reality must take precedence over public relations, for nature cannot be fooled. - R P Feynmann

@users =  sort { $::a->{count} <=> $::b->{count} }  @users;
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my $a = 'john';
@users =  sort { $a->{count} > $b->{count} }  @users;
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users is a hash, and we know this because of the brackets following $users: $users{$value}->{count}, as demonstrated in the original question.

Based on what the OP told us, the structure could minimally look something like this:

%users = (
    john  => { count => 1 },
    frank => { count => 3 }
);
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If the goal is to sort the list of users based on the value of count, you're not really doing that. You're jumping to the assumption that the datastructure looks more like this:

@users = [
    { count => 1 },
    { count => 3 }
];
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In other words, your solution assumes that @users is an array of hashrefs, as opposed to being a hash at its top level. I don't see how you're deriving that assumption based on the original question.

What your solution accomplishes is to sort the array, @users, based on $users->{count}, not based on $users{$value}->{count}.

Another perplexing topic covered in your reply is that of using '>' as the comparison operator within a sort routine. sort expects a comparison that returns either -1, 0, or 1. cmp does this, as does <=>. But there is nothing that will cause the '>' (greater than) operator to return a -1. So you're correct that "this won't work", but it has nothing to do with your choice of $::a and $::b versus $a and $b, and everything to do with an inappropriate use of the '>' operator. sort always uses a package global version of $a and $b anyway.




Dave

use strict;
use warnings;
my $users = {};
$users->{fred}->{count}=1;
$users->{sarah}->{count}=2;
$users->{arthur}->{count}=1;
$users->{benjamin}->{count}=10;
$users->{phil}->{count}=3;
$users->{betty}->{count}=4;
$users->{ilsa}->{count}=5;

print $_ . "\n" for sort { $users->{$a}->{count} <=> $users->{$b}->{co
+unt} } keys %$users;
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UPDATE: Or, you could use a hash, like the op did:

use strict;
use warnings;
my %users;
$users{fred}->{count}=1;
$users{sarah}->{count}=2;
$users{arthur}->{count}=1;
$users{benjamin}->{count}=10;
$users{phil}->{count}=3;
$users{betty}->{count}=4;
$users{ilsa}->{count}=5;

print $_ . "\n" for sort { $users{$a}->{count} <=> $users{$b}->{count}
+ } keys %users;
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for my $user ((sort { $users{$b}->{count} <=> $users{$a}->{count} } ke
+ys %users) {
   blah blah...
}
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Global symbol "$user" requires explicit package name at
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my @sorted_users = sort { $users{$b}->{count} <=>
    $users{$a}->{count} } keys %users;
for (@sorted_users) {
    blah... blahh
}
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Does the symbol $user appear somewhere outside of the loop? Its scope is constrained to inside the loop. If you try to access $user outside the loop you'll get a fatal error under strictures.

Also check to ensure that the line preceeding your for loop isn't missing a semicolon at the end of the line. ;)




Dave

Here is the entire loop:

for my $location (sort {$a cmp $b} keys %user_locations) {

    $worksheet->write($row++, $col, "$location  ($user_locations{$loca
+tion})", $format);

    my @sorted_users = sort { $users{$b}->{count} <=>
        $users{$a}->{count} } keys %users;

    for my $user (for my $user ((sort { $users{$b}->{count} <=> $users
+{$a}->{count} } keys %users) {

        next unless $users{$user}->{suffix} eq $user_locations{$locati
+on};

        for (my $tmp=1; $tmp <= keys %sheet_layout; $tmp++) {

            $worksheet->write($row, $col++, $users{$user}->{$sheet_lay
+out{$tmp}->[0]}, $format_r);
        }
        $col = 0;
        $row++;
    }
    $row++;
}
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I can't find the problem right now, maybe I should stare at it a while longer...