Wednesday, August 5, 2009

RE: Pass data to jquery

Here is the controller function

function add($showcase_id = null)
{
if (!empty($this->data)) {
$this->data['Entry']['id'] = $this->Entry->generate11Key();
$this->data['Entry']['showcase_id'] = $showcase_id;
if ($this->Entry->save($this->data)) {
$image_path =
$this->Image->upload_image_and_thumbnail($this->data, 'image', '640', '360',
'177', '100', 'entries', $this->Auth->user('slug'),
$this->data['Entry']['id']);
if (isset($image_path)) {
$this->Entry->saveField('image', $image_path);
$this->set('newImage', $image_path);
}
$this->Session->setFlash(__('New entry saved!.', true));
//$this->redirect(array('controller' => 'showcases',
'action' => 'view', $showcase_id));
} else {
$this->Session->setFlash(__('Error message. Please, try
again.', true));
}
}
$this->set('showcase', $showcase_id);
$this->set('entry', $this->data['Entry']['id']);

}

The response after upload shows this snip where newImage = what I need to
send to the script

$___viewFn = "/app/views/entries/add.ctp"
$___dataForView = array(
"slug" => "austinmaharaj",
"newImage" => "joesmith_25f31f1f34c.jpg",
"showcase" => "6e87ea500fe",
"entry" => "25f31f1f34c"
)

-----Original Message-----
From: brian [mailto:bally.zijn@gmail.com]
Sent: August-05-09 11:44 AM
To: cake-php@googlegroups.com
Subject: Re: Pass data to jquery


You haven't shown how/what your controller is sending back to the client.
Why does the response contain the SQL query? In any case, just send the new
filename back. Use JSON, for example. Have a look at the
json_encode() function.

On Wed, Aug 5, 2009 at 9:27 AM, Dave Maharaj ::
WidePixels.com<dave@widepixels.com> wrote:
> I am uloading a file to the server using an ajax upload script. Now
> everything is working fine except I cant figure out how to get the
> name of the file uploaded back to the page after uploading.
>
> Using Valums ajax upload script
>
> The file that gets uploaded say 03.jpg gets renamed to
> $user_$unique_id so it turns into myname_12565895.jpg on the server
>
> I used
> alert(response);  in the script after succesful upload and see
>
> UPDATE `entries` SET `image` = 'joekool_b9b071c1e64.jpg'  WHERE
> `entries`.`id` = 'b9b071c1e64'
>
> How can I grab the image` = 'joekool_b9b071c1e64.jpg'  and pass it
> back to the script rather than the 03.jpg. After upload it says 03.jpg
> uploaded succesfully but i need the server converted name so I can
> display the image instead of the name.
>
> Ideas?
>
> Dave
> >
>

--~--~---------~--~----~------------~-------~--~----~
You received this message because you are subscribed to the Google Groups "CakePHP" group.
To post to this group, send email to cake-php@googlegroups.com
To unsubscribe from this group, send email to cake-php+unsubscribe@googlegroups.com
For more options, visit this group at http://groups.google.com/group/cake-php?hl=en
-~----------~----~----~----~------~----~------~--~---

No comments: