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Using as_url

Jul 19, 2007, 1:01 PM

Eraser

Enthusiast (619 posts)

Jul 19, 2007, 1:01 PM

Post #1 of 2

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Using as_url

sub {
my $tags = shift;
my $lnkid = $tags->{'ID'};
my $db = $DB->table ('Category','CatLinks');
my $catname = $db->select ( { LinkID => $lnkid }, ['Full_Name'] )->fetchrow;
return $CFG->{build_root_url} . "/" . $catname . "/";
}

That returns a full category name from a link ID. The only issue is the output is like:

site.com/Cars/Ford Cruger/

rather than a correct category output:

site.com/cars/ford-escort/

This code I believe gives the correct result, but I'd be damed if I can integrate this right!

return $CFG->{build_root_url} . "/" . $cat_table->as_url($cat->{Full_Name}) . "/";

Any help really appreciated.

~ ERASER

Free JavaScripts @ Insight Eye

Jul 20, 2007, 2:02 AM

Andy

Veteran / Moderator (18441 posts)

Jul 20, 2007, 2:02 AM

Post #2 of 2

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Re: [Eraser] Using as_url In reply to

Hi,

I believe tihs should work:

Code:
sub { 
   my $tags = shift; 
   my $lnkid = $tags->{'ID'}; 
   my $db = $DB->table ('Category','CatLinks'); 
   my $catname = $db->select ( { LinkID => $lnkid }, ['Full_Name'] )->fetchrow; 
   return $CFG->{build_root_url} . "/" . $DB->table ('Category')->as_url( { Full_Name => $catname } ) . "/"; 
}

If not, it may be you need to get the CategoryID, and then do:

Code:
   return $CFG->{build_root_url} . "/" . $DB->table ('Category')->as_url( { ID => $catid } ) . "/";

Hope that helps.

Cheers

Andy (mod)
andy@ultranerds.co.uk

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