Django get file url - django

Question

I want to show the URL, filename and size of the uploaded file on my site.
I tried object.file.url but it didn't work.
models.py:
file = models.FileField(upload_to="media/",null=True,blank=True, verbose_name="Files", validators=[
FileExtensionValidator(allowed_extensions=['pdf', 'docx', 'doc', 'xlsx', 'xls', 'png','jpg','jpeg'])])
And tried this but not working:
in models.py:
#property
def file_url(self):
if self.file and hasattr(self.file, 'url'):
return self.file.url

Answer 1

You didn't give enough information to solve this issue . Where is your views.py and templates/filename.html file code ?
Since you didn't give enough information so I try to answer your question by using another example.
Suppose your models.py file is look like in the below :
models.py
class Pic(models.Model):
name = models.CharField(max_length=255)
photo = models.FileField(upload_to="data/media/%Y/%m/%d")
def __str__(self):
return self.url
def path(self):
return self.url
And also your views.py is look like in the below:
views.py
def upload_pic(request):
documents = Pic.objects.all()
import uuid
name = uuid.uuid4().hex[:6].upper()
if request.method == 'POST':
form = DocumentForm(request.POST, request.FILES)
if form.is_valid():
print(request.FILES)
newdoc = Pic(name=name, photo=request.FILES['docfile'])
newdoc.save()
return render(request, 'clinic/list.html', {'documents': documents, 'form': form, 'msg': 'success'})
else:
form = DocumentForm() # A empty, unbound form
return render(request, 'clinic/list.html', {'documents': documents, 'form': form})
In that case , you can show your uploaded file's url by using {{ document.photo.url }} that line of code in your template file . just check out the below example :
<!-- List of uploaded documents -->
{% if documents %}
<ul>
{% for document in documents %}
<li>{{ document.photo.name }}</li>
{% endfor %}
</ul>
{% else %}
<p>No documents.</p>
{% endif %}