Hi Limbic~Region,

Thanks for the reply. I tested this code earlier also, but didnt get answer properly. ie there is a bug in logic where we assigns to the hash. The output we get is

1	Virginia	Norfolk
2	Virginia	Chesapeake
3	Virginia	Virginia Beach
6	Virginia	Falls Church
8	Washington	Seatle
9	Washington	Spokane
4	Indiana	Evansville
5	Indiana	Fort Wayne
7	Indiana	Indianapolis

This is NOT the o/p I need. So I am still trying to get it correct. (didnt gaveup yet)

"It worries me that you are testing code you don't understand"

Its another way how I am learn. I try to change the code and learn all possibilities :)

Thanks for your help & reply

-- VC

vcTheGuru,
There isn't a bug in the logic. The bugs were in the code, which was untested. To show that the logic is correct. Here is the code working on your data set with very minor bug fixes:

#!/usr/bin/perl
use strict;
use warnings;

my %data;
while (<DATA>) {
    chomp;
    my @col = split /\|/;
    if (exists $data{$col[1]}) {
        $data{$col[1]}{min} = $col[0] if $col[0] < $data{$col[1]}{min}
+;
        push @{$data{$col[1]}{entry}}, \@col;
    }
    else {
        $data{$col[1]}{min} = $col[0];
        $data{$col[1]}{entry} = [\@col];
    }
}
for my $state (sort {$data{$a}{min} <=> $data{$b}{min}} keys %data) {
    for my $entry (sort {$a->[0] <=> $b->[0]} @{$data{$state}{entry}})
+ {
        print "@$entry\n";
    }
}


__DATA__
1|Virginia|Norfolk
2|Virginia|Chesapeake
3|Virginia|Virginia Beach
4|Indiana|Evansville
5|Indiana|Fort Wayne
6|Virginia|Falls Church
7|Indiana|Indianapolis
8|Washington|Seatle
9|Washington|Spokane
[download]

I see that you have a SQL solution not requiring perl.

Cheers - L~R