Wednesday, August 5, 2009

Re: Pass data to jquery

I'm still not getting it. If there's no view, how is it that
javascript receives that block you posted. And why do you have
$this->set('newImage', $image_path); if there's no view to set it for?

On Wed, Aug 5, 2009 at 12:47 PM, Dave Maharaj ::
WidePixels.com<dave@widepixels.com> wrote:
>
> That's what shows up in the javascript
>
> alert(response);
>
>
> onComplete: function(file, response){
>                        button.text('Upload');
>                        alert(response);
>                        $('.saving').hide();
>                                        $('#example1 .text').text('Upload
> Complete');
>                        // enable upload button
>                        this.enable();
>
>                        // add file to the list
>                        //$('<li></li>').appendTo('#example1
> .files').text(file);
>
>                                        $('<li></li>').appendTo('#example1
> .files').html('<img src="/img/entries/thumb/" alt="" /><br />');
>
>
>                }
>
> There is no view file for the returned image, just a button on the page
> "Upload" which when clicked upens the file browser on users
> computer...select image upload and when complete the alert response.
>
> Dave
>
> -----Original Message-----
> From: brian [mailto:bally.zijn@gmail.com]
> Sent: August-05-09 1:51 PM
> To: cake-php@googlegroups.com
> Subject: Re: Pass data to jquery
>
>
> I'm not sure that I understand what that response is. Is that your view
> file? How are you passing that back to javascript?
>
> On Wed, Aug 5, 2009 at 10:23 AM, Dave Maharaj ::
> WidePixels.com<dave@widepixels.com> wrote:
>>
>> Here is the controller function
>>
>> function add($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" =&gt; "austinmaharaj",
>>        "newImage" =&gt; "joesmith_25f31f1f34c.jpg",
>>        "showcase" =&gt; "6e87ea500fe",
>>        "entry" =&gt; "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
>>> >
>>>
>>
>>
>>
>> >
>>
>
>
>
> >
>

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