That explains your problem better :).
Do you need all the data from that query? Or only photo_id? In case
you just need the photo_id you can do:
$photoIds = $this->Hotel->findAll("hotel_id = '$hotel_id'", "DISTINCT
photo_id");
Using GROUP BY when there is no agregation function in SELECT clause
somehow works in MySQL (I didn't believe that until I checked this
myself), but for example, in MSSQL this raises an error. I think that
MySQL is here not compliant with the standard.
On 21 Sty, 16:01, cronet <[EMAIL PROTECTED]> wrote:
> Thanks for your replies!
>
> @Claudia, I will try this!
>
> @Marcin
>
> I think an order by is not enough in this case.
>
> There are multiple entries with the same photo_id, but different
> descriptiontype_id's ... But at the end they are the same photos. Only
> in different categories (descriptiontypes). This structure is fixed by
> an webservice, so no possibility to change that fact.
>
> But i want to show them only once (it doesn't matter which category).
>
> So an simple ORDER would not prevent showing the same photo.
>
> I would use an DISTINCT, but i neet other informations too, so google
> said (:) I should rather use a GROUP BY clause...
>
> On 21 Jan., 12:03, Marcin Jaworski <[EMAIL PROTECTED]> wrote:
>
> > I don't see a reason for using grouping in this query. 'Group by' is
> > only needed if you want to use some aggregation functions like SUM,
> > AVG, MAX, MIN etc. If you don't use them, you don't need 'group by' in
> > your query. I think that use of 'order by' would be enough in your
> > situation, so you need to add an order value to findAll to order it by
> > photo_id.
>
> > On 20 Sty, 23:10, cronet <[EMAIL PROTECTED]> wrote:
>
> > > Hi,
>
> > > I'm siting here for about 2 hours and get it not work. I need a simple
> > > group by in an association.
>
> > > This is in my controller:
>
> > > $this->Hotel->bindModel(
> > > array('hasMany' =>
> > > array(
> > > 'HotelDescriptionTranslation' => array(
> > > 'className' =>
> > > 'HotelDescriptionTranslation',
> > > 'foreignKey' =>
> > > 'hotel_id',
> > > 'conditions' =>
> > > $cond1
> > >
> > > ),
> > > 'HotelDescriptionPhoto' => array(
> > > 'className' =>
> > > 'HotelDescriptionPhoto',
> > > 'foreignKey' =>
> > > 'hotel_id',
> > > 'conditions' =>
> > > $cond2
> > > )
> > > )
> > > ), FALSE
> > > );
> > > $hotel = $this->Hotel->findByHotelId($hotel_id);
>
> > > Results in this query:
>
> > > SELECT `HotelDescriptionPhoto`.`id`,
> > > `HotelDescriptionPhoto`.`descriptiontype_id`,
> > > `HotelDescriptionPhoto`.`hotel_id`,
> > > `HotelDescriptionPhoto`.`photo_id`,
> > > `HotelDescriptionPhoto`.`url_max300`,
> > > `HotelDescriptionPhoto`.`url_original`,
> > > `HotelDescriptionPhoto`.`url_square60`,
> > > `HotelDescriptionPhoto`.`created`, `HotelDescriptionPhoto`.`modified`
> > > FROM `hotel_description_photos` AS `HotelDescriptionPhoto` WHERE
> > > (descriptiontype_id = '5' OR descriptiontype_id = '10') AND
> > > `HotelDescriptionPhoto`.`hotel_id` IN (60160)
>
> > > But I need a "GROUP BY `HotelDescriptionPhoto`.`photo_id`" inside the
> > > query...
>
> > > Is there an easy(?!) way to achieve this? (I would not want to create
> > > a custom query)...
>
> > > Regards,
> > > Alexander
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