[Answer]-Django-Provide a download from a function defined in a model

by

1👍

✅

First point: a response is not an url. What you want in your template are urls, not responses. Second point: generating a reponse is the responsability of a view, not of a model. Side note : you should respect Python’s coding conventions (cf pep08)

The RightWay(tm) to organize your code would be:

# myapp/models.py
class TestResultFile(models.Model):
    user=models.ForeignKey(User)
    system=models.ForeignKey(System)
    test_id=models.ForeignKey(Detail)
    path=models.CharField(max_length=300)
    class Meta:
        verbose_name="Test result file"
        verbose_name_plural="Test result files"

# myapp/views.py
def download_file(request, file_id):
    testfile = get_object_or_404(TestResultFile, pk=file_id)
    wrapper = FileWrapper(open(testfile.path, "r" ))
    response=HttpResponse(wrapper, content_type="text/plain")
    response['Content-Disposition'] ='attachment; filename="results.txt"'
    return response 

# myapp/urls.py
urlpatterns = patterns('',
    url(r'^download/(?P<file_id>\d+)/?', 'views.download_file', 'myapp_download_file'),
    # ...
    )

# myapp/templates/myapp/template.html
<ul>
{% for at in attempts %}
  <li>
      System Name: <em>"{{ at.system}}"</em>, 
      download file: <a href="{% url 'download_file' at.pk %}">here</a> 
  </li>
 {% endfor %}
 </ul>
👤bruno desthuilliers

0👍

Maybe you can simply use somthing like this:

class userfile(model.Model):
    user=models.ForeignKey(User)
    file = models.FileField(_('file'), upload_to='userfile/', blank=False)

    def __unicode__(self):
        return "%s file" % user

and in you template:

 {% if user.userfile_set.count > 0 %}
 <ul>
 {% for file in user.userfile_set.all %}
 <li>File: <a href="{{MEDIA_URL}}{{file.file}}">{{file}} dowload it</a></li>
 {% endfor %}
 </ul>
 {% else %}
 You don't have any file
 {% endif %}

I hop it can help you.

👤Krozark

Leave a comment