To force an update.. just set the id of the Model
example:
$this->ContentBlock->id = 3;
$this->ContentBlock->save( $quote );
On Apr 25, 11:21 am, Mike Digital Egg <[EMAIL PROTECTED]>
wrote:
> Hi,
>
> Thanks, that worked, didn't know about that function.
>
> Do you know if there is an equivalent function to force an UPDATE.
>
> Cheers
>
> Mike
>
> On Apr 25, 6:34 pm, rtconner <[EMAIL PROTECTED]> wrote:
>
> > You should make a a practice to call $this->Model->create() before
> > every insert.
>
> > So in this example..
>
> > $this->ContentBlock->create();
> > $this->ContentBlock->save( $quote );
> > ...
>
> > $this->ContentBlock->create();
> > $this->ContentBlock->save( $header );
>
> > On Apr 25, 10:11 am, Mike Digital Egg <[EMAIL PROTECTED]>
> > wrote:
>
> > > Hi,
>
> > > I am having a problem trying to save 2 different values to a single
> > > table, below is the code:
>
> > > if ( $_POST["data"]["Block"][1] ) {
> > > $quote = array(
> > > 'content_id' => $id,
> > > 'block_id' => $_POST["data"]["Block"][1]["id"]
> > > );
> > > $this->ContentBlock->save( $quote );}
>
> > > if ( $_POST["data"]["Block"][2] ) {
> > > $header = array(
> > > 'content_id' => $id,
> > > 'block_id' => $_POST["data"]["Block"][2]["id"]
> > > );
> > > $this->ContentBlock->save( $header );
>
> > > }
>
> > > INSERT INTO `content_blocks` (`content_id`,`block_id`) VALUES
> > > ('22','1')
> > > SELECT COUNT(*) AS count FROM `content_blocks` WHERE (`id` = '59')
> > > UPDATE `content_blocks` SET `content_id` = '22',`block_id` = '8' WHERE
> > > `id` = '59'
>
> > > It is inserting the first request but then overwriting it with the
> > > second, rather than adding a new entry.
>
> > > Is there a better way of doing this.
>
> > > Thanks
>
> > > Mike
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