Re: One model, multiple pagination in a single view with conditions?
Gah that'd be right. First time jumping into CakePHP, the project I'm
working on needs pagination of HABTM model (which was difficult), and
multiple pagination of one model on the same view (which is impossible). I
could have sorted these problems in no time with pure PHP.
Oh well, I ended up going with a simple iframe to have multiple pagination
on the same page. A cheap hack, but it works.
Cheers
On Thursday, June 28, 2012 4:00:59 AM UTC+10, Mark Wratten wrote:
You can only paginate one way at a time, so presumably you can set the
associations dynamically depending on the parameters from jQuery. Though
separate models might be cleaner as you can hard-code the associations.
Mark
On Monday, June 25, 2012 11:41:16 PM UTC-4, 42startups wrote:
As the title suggests, is there a simple way to do this?
Or should I create a separate model?
Here's the code:
$this-paginate = array('limit'=30);
$this-Hashtag-Pin-bindModel(array('hasOne' = array('HashtagsPin')),
false);
$this-set('hashtag', $this-paginate('Pin',
array('HashtagsPin.hashtag_id' = $id)));
$this-set('thought', $this-paginate('Pin',
array('HashtagsPin.hashtag_id' = $id,'Pin.url'=null)));
So I need to paginate 'hashtag' and 'thought' separately. I'll be
attaching infinite scroll jquery to both.
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Re: One model, multiple pagination in a single view with conditions?
You can only paginate one way at a time, so presumably you can set the
associations dynamically depending on the parameters from jQuery. Though
separate models might be cleaner as you can hard-code the associations.
Mark
On Monday, June 25, 2012 11:41:16 PM UTC-4, 42startups wrote:
As the title suggests, is there a simple way to do this?
Or should I create a separate model?
Here's the code:
$this-paginate = array('limit'=30);
$this-Hashtag-Pin-bindModel(array('hasOne' = array('HashtagsPin')),
false);
$this-set('hashtag', $this-paginate('Pin',
array('HashtagsPin.hashtag_id' = $id)));
$this-set('thought', $this-paginate('Pin',
array('HashtagsPin.hashtag_id' = $id,'Pin.url'=null)));
So I need to paginate 'hashtag' and 'thought' separately. I'll be
attaching infinite scroll jquery to both.
--
Our newest site for the community: CakePHP Video Tutorials
http://tv.cakephp.org
Check out the new CakePHP Questions site http://ask.cakephp.org and help others
with their CakePHP related questions.
To unsubscribe from this group, send email to
cake-php+unsubscr...@googlegroups.com For more options, visit this group at
http://groups.google.com/group/cake-php
Re: One model, multiple pagination in a single view with conditions?
I don't think it's possible with cake.
Le mardi 26 juin 2012 05:41:16 UTC+2, 42startups a écrit :
As the title suggests, is there a simple way to do this?
Or should I create a separate model?
Here's the code:
$this-paginate = array('limit'=30);
$this-Hashtag-Pin-bindModel(array('hasOne' = array('HashtagsPin')),
false);
$this-set('hashtag', $this-paginate('Pin',
array('HashtagsPin.hashtag_id' = $id)));
$this-set('thought', $this-paginate('Pin',
array('HashtagsPin.hashtag_id' = $id,'Pin.url'=null)));
So I need to paginate 'hashtag' and 'thought' separately. I'll be
attaching infinite scroll jquery to both.
Any ideas?
Cheers
--
Our newest site for the community: CakePHP Video Tutorials
http://tv.cakephp.org
Check out the new CakePHP Questions site http://ask.cakephp.org and help others
with their CakePHP related questions.
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cake-php+unsubscr...@googlegroups.com For more options, visit this group at
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