CODEWITHSUNDEEP
pythondjangojquery
Asma Gheisari
Asma Gheisari

Reputation: 6254

getting data of submitted file using jquery ajax and handling It with ModelForm

I'm going to edit an ImageField using jquery ajax,after searching I found out I should use jquery form plugin,so I think using introduced method in this article is good : http://www.laurentluce.com/posts/upload-to-django-with-progress-bar-using-ajax-and-jquery/ But I wanna get data of POST image and give it to a model form to handle It,because It take care of every thing,even when the name of file is repeated,some thing like this:

<form id='form_upload' action="." method="POST" enctype="multipart/form-data">
    <input type='file' id='id_HeadImage' />
</form>
<script typr="text/javascript">    
var options = {
      dataType: 'xml',
      url: '{% url DrHub.views.editNews param1,param2 %}',
      beforeSubmit: showRequest,
      success: showResponse
    }
    $('#form_upload').ajaxSubmit(options);
</script>

and in server side :

if ('id_HeadImage' in request.FILES) and (request.FILES['id_HeadImage']):
    gForm=GalleryForm(request.FILES['id_HeadImage'],instance=newsInstance.gallery_ptr)
    if gForm.is_valid():
       gForm.save()

as U can see I'm going to edit ImageField of a model named Gallery but this way I just get the file name.

How can I do this?

EDIT

this is Gallery Model:

class Gallery(models.Model):
   HeadImage = models.ImageField(upload_to="gallery",blank=True,null=True)

EDIT

Request.FILES is empty and ajax just post file name,not data!

Upvotes: 0

Views: 372

Answers (1)

jpic
jpic

Reputation: 33410

  1. Set the name attribute of your file input

  2. Instanciate the form with request.POST and request.FILES as such: gForm=GalleryForm(request.POST, request.FILES, instance=newsInstance.gallery_ptr)

Also this:

if ('id_HeadImage' in request.FILES) and (request.FILES['id_HeadImage']):

Could be more pythonic:

if 'id_HeadImage' in request.FILES.keys():

Upvotes: 1

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