Perl排序哈希，如何按$ hash排序{$ key} - > {secondkey}

Question

I have a Perl function:

my %d;

$d{"aaaa"}->{t1} = "9:49";
$d{"bbbb"}->{t1} = "9:30";

foreach my $k (sort { ($d{$a}->{t1}) <=> ($d{$b}->{t1}) } keys %d)
{
    print "$k:  $d{$k}->{t1}\n";
}

I want to sort by t1, so 9:30 before 9:49 and I want to get the result:

bbbb:  9:30
aaaa:  9:49

but the result is not suitable.

It seems like the result is random?

C:\tmp>a.pl
bbbb:  9:30
aaaa:  9:49

C:\tmp>a.pl
bbbb:  9:30
aaaa:  9:49

C:\tmp>a.pl
bbbb:  9:30
aaaa:  9:49

C:\tmp>a.pl
aaaa:  9:49
bbbb:  9:30

C:\tmp>a.pl
bbbb:  9:30
aaaa:  9:49

C:\tmp>a.pl
bbbb:  9:30
aaaa:  9:49

C:\tmp>a.pl
bbbb:  9:30
aaaa:  9:49

C:\tmp>a.pl
aaaa:  9:49
bbbb:  9:30

Answer 1

You need to use cmp instead of <=> since you are comparing strings. The comments are correct and we need to take into consideration 10+ hours. You need to use sprintf to add leading zero when hours are less than 10 to have strings sorted correctly.

foreach my $k (sort { sprintf("%05s", ($d{$a}->{t1})) cmp sprintf("%05s", ($d{$b}->{t1})) } keys %d) {

Answer 2

<=> is for comparing numbers, but your times have a colon, which makes them strings instead of numbers. One workaround is to just remove the colon, so that <=> can operate on them in number context.

use v5.10;

say "$_: $d{$_}->{t1}" for sort { $d{$a}->{t1} =~ s/://r <=> $d{$b}->{t1} =~ s/://r } keys %d;

The r modifier on the substitution means return the new value without altering the old value.