Re: Search an array ref of array refs

Thanks to the good 'ol folks at google and Perl Limericks:

my $domain_ip;    
map { if($_->[0] == $domainid){ $domain_ip = @$_->[2] } } @$table;
[download]

Tx all! :)

Comment on Re: Search an array ref of array refs Download Code

Replies are listed 'Best First'.
Re: Re: Search an array ref of array refs by blakem (Monsignor) on Nov 01, 2001 at 08:49 UTC
I think grep makes more sense in this case. Assuming you know there is only one result per $id, and don't mind checking the entire datastructure, this should work: `my $domain_ip = (grep {$_->[0] == $domainid} @$table)[0]->[2];` [download] As in the folloing example: `#!/usr/bin/perl -wT use strict; my $table = [ [7, 'tigers', 600], [8, 'elephants', 250], [9, 'lions', 500], ]; my $id = 8; my $fieldindex = 1; my $animal = (grep {$_->[0] == $id} @$table)[0]->[$fieldindex]; print "id=$id fieldindex=$fieldindex => animal=$animal\n"; =OUTPUT id=8 fieldindex=1 => animal=elephants` [download] -Blake	[reply] [d/l] [select]

Replies are listed 'Best First'.

Re: Re: Search an array ref of array refs
by blakem (Monsignor) on Nov 01, 2001 at 08:49 UTC

my $domain_ip = (grep {$_->[0] == $domainid} @$table)[0]->[2];
[download]

#!/usr/bin/perl -wT
use strict;

my $table =  [ [7, 'tigers',    600],
               [8, 'elephants', 250],
               [9, 'lions',     500], ];

my $id         = 8;
my $fieldindex = 1;

my $animal = (grep {$_->[0] == $id} @$table)[0]->[$fieldindex];

print "id=$id fieldindex=$fieldindex => animal=$animal\n";

=OUTPUT
id=8 fieldindex=1 => animal=elephants
[download]

-Blake

[reply]
[d/l]
[select]