Re: Displaying a document

I think my code is more or less what you said, i made the change in the template to include a link to the uploaded file using the Media_URL variable, but it didn't work. It gave me a page not found error. I don't know if i have to include a URL, the documentation doesn't mention anything. Below is my code if you can please check it out and tell me what i am missing.

This is my URL Conf:

 url(r'^candidatelist/'
        r'(?P<year>\d{4})/'
        r'(?P<month>\d{1,2})/'
        r'(?P<day>\d{1,2})/'
        r'(?P<slug>[-\w]*)/'
        r'update/'
        r'resume/'
        r'(?P<resumeFile>)/$',
        Candidate_Update.as_view(),
        name = 'resume_view')

This is my model:

class Candidate(models.Model):
   
    resumeFile = models.FileField(upload_to="resume/")
   

This is my template:

        {% for candidate in candidatelist %}
        <tr>
            <td><a href="{{candidate.get_update_url}}">{{candidate.lastname}}</a>
            </td>
            <td>{{candidate.firstname}}</td>
            <td>{{candidate.career}}</td>
            <td>{{candidate.email}}</td>
            <td><a href="{{ MEDIA_URL }}">{{candidate.resumeFile}}"</a></td>
        </tr>
        {% endfor %}
        </tbody>
    </table>
   
   
</div>

{% endblock %}

and this is my view:

def CandidateList(request):
    candidatelist = Candidate.objects.all()
    return render(request, 'recruitment/candidatelist.html', {'candidatelist': candidatelist})

           
class Candidate_Detail(ObjectUpdateMixin, View):
    form_class = CandidateForm
    model = Candidate
    template_name = 'recruitment/candidate_detail.html'


class Candidate_Update(ObjectUpdateMixin, View):
    form_class = CandidateForm
    model = Candidate
    template_name = 'recruitment/candidate_update.html'

with the mixin:

class ObjectCreateMixin:
    form_class = None
    template_name = ' '

    def get(self, request):
        return render(
                      request,
                      self.template_name,
                      {'form': self.form_class()}
                      )
   
    def post(self, request):
        bound_form = self.form_class(request.POST, request.FILES)
        if bound_form.is_valid():
            new_object = bound_form.save()
            return redirect(new_object)
        else:
            return render(
                          request,
                          self.template_name,
                          {'form' : bound_form})
           

class ObjectUpdateMixin:
    form_class = None
    model = None
    template_name = ' '
   
    def get(self, request, slug, **kwargs):
        obj = get_object_or_404(self.model,
                                 slug__iexact=slug
                                          )
       
        context = {'form' : self.form_class(instance=obj),
                   self.model.__name__.lower() : obj,}
        return render(
                      request,
                      self.template_name,
                      context
                      )
   
    def post(self, request, slug, **kwargs):
        obj = get_object_or_404(self.model, slug=slug)
        bound_form = self.form_class(request.POST, request.FILES, instance=obj)
        if bound_form.is_valid():
            new_object = bound_form.save()
            return redirect(new_object)
        else:
            context = {
                       'form' : bound_form,
                       self.model.__name__.lower() : obj,
                       }
           
            return render(
                          request,
                          self.template_name,
                          context)

On Saturday, May 21, 2016 at 11:07:11 AM UTC-7, luisza14 wrote:

Do you have a model  with a fileField?

I think you are wrong passing file to template, if  you have a model with filefield you can use media url something like

class Mymodel(Model):

      myfile = models.FileField(upload_to="my_file_path_in_media")

so you can use a form

class Myform(forms.ModelForm):

     class Meta:

             model=Mymodel

             fields = '__all__'

so in the view you have

if request.method=='POST':

       form = MyForm(request.POST, request.FILE)

       if form.is_valid():

            instance=form.save()

so you can pass instance in template context

       render(request, 'mytemplate.html', {'instance': instance}

in your template you can put a link or whatever you want.

<a href="{{instance.myfile.url}}"> download</a>

take a look https://docs.djangoproject.com/en/1.9/topics/http/file-uploads/ form more info

2016-05-21 10:43 GMT-06:00 Wilfredo Rivera <wriv...@gmail.com>:

Hello:

I want to display an uploaded file in the browser. My website save the file in the database, but i don't have idea of how can the website display it once the user click on the link. I am new to django and programming in general so i am learning on a trial and error basis.

The view that handles this is the following:

def File_Open(request):
    if (request == request.FILES):
       file = request.FILES.open()
       return render(request, 'recruitment/open_file.html', {file : 'file'})

and the url is this:

url(r'^candidatelist/'
        r'resume-view/'
        r'(?P<resumeFile>/$',
        views.File_Open,
        name = 'resume_view'

I have search for how can i do this, but haven't found a satisfying answer. Please i will appreciate the help.

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