Django-Provide a download from a function defined in a model - django

Question

I am trying to implement a view, where a logged in user that has uploaded a file can download his file, but only his, not other users files, so I don't create an id url based on the pk of the file . As a result in the views I query the test_result_file table and I filter it for a specific user. I think that I can do what I want by writing a function in my model:
class test_result_file(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"
def get_self(self):
path=self.path
wrapper = FileWrapper(open( path, "r" ))
response=HttpResponse(wrapper, content_type="text/plain")
response['Content-Disposition'] ='attachment; filename="results.txt"'
return response
However, in the template, when I call:
<ul>
{% for at in attempts %}
<li>System Name: <em>"{{ at.system}}"</em>, download file: here </li>
{% endfor %}</ul>
the download is not provided and instead the browser tries to open a url with all the parameters of response and fails.
Am I losing something? Is this feasible with that function?

Answer 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: here
</li>
{% endfor %}
</ul>

Answer 2

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: {{file}} dowload it</li>
{% endfor %}
</ul>
{% else %}
You don't have any file
{% endif %}
I hop it can help you.