$data = $this->Model->find etc...

unset($data['Model']['id'];

$this->data = $data;


On 14 Aug 2013, at 19:40, Anja Liebermann <[email protected]> wrote:

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> Hi David,
> 
> 
> this is how I would do it:
> 
> copy your edit method and rename it to "copy". Like with edit you fetch
> the data which should be cloned and show then in a view "copy.ctp" which
> has a form like for adding a new record.
> Then when the user submits you call your copy method again in the "hey I
> have post data" fork and save it as a new record by unsetting a probably
> exiting id.
> 
> That should do it.
> 
> HTH!
> Anja
> 
> 
> 
> Am 14.08.2013 20:22, schrieb David Carr:
>> Here's what I am trying to accomplish:
>> 
>> A user fills out a form, saves a record.  At some later date they wish to 
>> "clone" this record, but may want to make a few tweaks.  This "clone" 
>> functionality should direct them to a form that is pre-filled with the 
>> previous record's data so that they can review it, edit as needed, and 
>> submit it as a new record.
>> 
>> What I'm trying:
>> 
>> I've modified my add() function to accept a parameters:
>> 
>> function add($cloneid = NULL)
>> 
>> Then created a Clone link that sends them to <site>/model/add/<id>
>> 
>> Then, I get the data from that model:
>> 
>> $clone_source = $this->Model->findById($cloneid);
>> $this->data['Model']['field1'] = $clone_source['Model']['field1'];
>> 
>> and so on.  Based on Google searching and other posts, this should work. 
>> But what actually happens is that upon clicking the 'Clone' link, the user 
>> is directed and the form submits itself immediately (failing to save the 
>> record, since it fails validation) and the user never actually sees the 
>> form.
>> 
>> What am I doing wrong?  (Also I should note, there are relational models 
>> present, but I don't think this should be the cause of any problems...I 
>> hope).
>> 
> 
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