Primarily to populate other relationships. So, for instance.
I have a two classes called 'Template' and 'Template::Version'.
Template has a one to many relationship with Template::Version
(Template::Version has a foreign key into Template),
as well as a many to one relationship with Template::Version through
'use_version_id', (which usually contains the most recent template version, but
not always).
I would like to be able to do....
use Template;
my $template = Template->new( name => 'blah blah!' );
my ($newversion) = $template->add_versions({value => 'xxyy', digest =>
'dasfwgwqw'});
$template->use_version($newversion);
$template->save;
Right now, I have to explicitly know the class and all attributes of
Template::Version to do this, i.e.
use Template;
use Template::Version;
my $template = Template->new( name => 'blah blah!' );
my $newversion = Template::Version->new( template => $template, value =>
'xxyy', digest => 'dasfwgwgw' );
$template->add_versions($newversion);
$template->use_version($newversion);
$template->save;
It just feels clumsy, and requires more hardcoded knowledge than I would like
at this level of abstraction.
Seems easy to correct as well?
I guess the question I should ask is:
Why NOT return the newly created objects?
On Wed, May 10, 2006 at 02:43:11PM -0400, John Siracusa wrote:
> On 5/10/06, Jesse Brown <[EMAIL PROTECTED]> wrote:
> >When adding objects to a relationship via $object->add_<relationship>({
> >blah
> >=> 1, etc}), I seem unable to get back the list of objects that were
> >created
> >and added to the relationship, which seems like it would be VERY
> >convenient?
> >
> >It also seems that the behaviour of: returning the number of items added in
> >scalar context, and a list of created objects when in list context would be
> >easy to add, and not break existing code.
>
> What do you plan to do with the objects returned, or with the count
> for that matter? Surely you know the count when you make the call
> because you know how many arguments you're passing. What's your usage
> scenario?
>
> -John
>
>
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