How can I upload multiple files to a model field? - django

Question

I want to upload multiple files through a ModelForm,with all files to be assigned to a file field of the Model.I have gone through the docs and I saw an example on it and I ve implemented it here but I can only get my form to pick multiple files but only one get saved and assigned to filesfield.Below are my codes
models.py
class Feed(models.Model):
user=models.ForeignKey(User,on_delete=models.CASCADE,related_name='feeds')
text=models.TextField(blank=False,max_length=500)
files = models.FileField(upload_to="files/%Y/%m/%d")
forms.py
class FeedForm(ModelForm):
class Meta:
model=Feed
fields=('text','auth','files')
widgets={"files":forms.FileInput(attrs={'id':'files','required':True,'multiple':True})}
and views.py
def post_feed(request):
form_class = FeedForm
if request.method == 'POST':
form = form_class(request.POST,request.FILES)
if form.is_valid():
feed = form.save(commit=False)
feed.user = User.objects.get(pk=1)
feed.pub_date=timezone.now()
#instance = Feed(files=request.FILES['files'])
# feed.files=request.FILES['files']
feed.save()
return redirect('home')
else:
form = form_class()
return render(request, 'post_feed.html', {'form': form,})
from django.views.generic.edit import FormView
from .forms import FeedForm
class FileFieldView(FormView):
form_class=FeedForm
template_name='post_feed.html'
'''success_url=??? #I dont know what to write here.I thought of putting this
render(request, 'post_feed.html', {'form': form,}) because I just want
to reload the page but it gave an error,so I removed it entirely.'''
def post_feed(self,request,*args,**kwargs):
form_class=self.get_form_class()
form=self.get_form(form_class)
filez=request.FILES.getlist('files')
if form.is_valid():
for f in filez:
f.save()
return self.form_valid(form)
else:
return self.form_invalid(form)
Kindly help me out,Thanks in advance.

Answer 1

You have to create a separate model for the files and connect them with a foreign key:
class Feed(models.Model):
user=models.ForeignKey(User, on_delete=models.CASCADE, related_name='feeds')
text=models.TextField(blank=False, max_length=500)
class FeedFile(models.Model):
file = models.FileField(upload_to="files/%Y/%m/%d")
feed = models.ForeignKey(Feed, on_delete=models.CASCADE, related_name='files')
I hope this helps.

Answer 2

Phew, it took me a whole day to figure out this. My goal was to assign multiple files to one instance of a class, like a Blog instance can have multiple Images. First things first, you cannot do this with one models.FileField inside a model (for example inside Blog class), because this field was not designed to save multiple files. So the solution is to create separate model for the files and connect them with One-to-Many Relationship (Foreign Key) as it was answered by #Carlos Mermingas.
Enough words, here is the code for the above situation:
# models.py
class Feed(models.Model):
user=models.ForeignKey(User, on_delete=models.CASCADE)
text=models.TextField(blank=False, max_length=500)
class FeedFile(models.Model):
file = models.FileField(upload_to="files/%Y/%m/%d")
feed = models.ForeignKey(Feed, on_delete=models.CASCADE)
# forms.py
...
from django.forms import ClearableFileInput
...
class FeedModelForm(forms.ModelForm):
class Meta:
model = Feed
fields = ['text']
class FileModelForm(forms.ModelForm):
class Meta:
model = FeedFile
fields = ['file']
widgets = {
'file': ClearableFileInput(attrs={'multiple': True}),
}
# widget is important to upload multiple files
# views.py
from .models import FeedFile
...
def create_to_feed(request):
user = request.user
if request.method == 'POST':
form = FeedModelForm(request.POST)
file_form = FileModelForm(request.POST, request.FILES)
files = request.FILES.getlist('file') #field name in model
if form.is_valid() and file_form.is_valid():
feed_instance = form.save(commit=False)
feed_instance.user = user
feed_instance.save()
for f in files:
file_instance = FeedFile(file=f, feed=feed_instance)
file_instance.save()
else:
form = FeedModelForm()
file_form = FileModelForm()
# the rest is the basic code: template_name, context, render etc.
# in your template.html <form> tag must include enctype="multipart/form-data"
Bonus: if you want to see uploaded files in admin panel, you can use InlineModelAdmin objects. Here is the code:
# admin.py of your app
from django.contrib import admin
from .models import Feed, FeedFile
class FeedFileInline(admin.TabularInline):
model = FeedFile
class FeedAdmin(admin.ModelAdmin):
inlines = [
FeedFileInline,
]
admin.site.register(Feed, FeedAdmin)
For the more details on file upload, Model Forms, how to include widget in Model Form

Answer 3

Would suggest using an M2M field from Feed model to FeedFile model.
Makes it all the more easier while querying for files of a particular Feed object, which i feel is also the most common usecase for Feed objects
class Feed(models.Model):
user=models.ForeignKey(User, on_delete=models.CASCADE, related_name='feeds')
text=models.TextField(blank=False, max_length=500)
files=models.ManyToManyField(FeedFile)
class FeedFile(models.Model):
file = models.FileField(upload_to="files/%Y/%m/%d")