Ohh thanks Tom
it helped !!
In views.py i changed ifile=request.POST.get('inputfile') to
ifile=request.FILES['inputfile']
and it worked....
Thanks a lot !!!
On Fri, Apr 17, 2015 at 8:59 PM, Tom Evans <tevans...@googlemail.com> wrote:
> On Fri, Apr 17, 2015 at 3:45 PM, livelikehimanshu12
> <livelikehimansh...@gmail.com> wrote:
> > Hello friends !!
> > So my project requires me to store images on blob service of windows
> azure
> > .I am using django in visual studio express . I even have an azure
> account..
> > So I wrote the code using a lot help from internet (I was total beginner
> to
> > django when I started this project ) . My code works perfectly fine for
> > django admin interface i.e. my images are getting stored on azure blobs
> > online and using the stored URL(in my models field) I am able to access
> it
> > also.
> > I followed this url https://pypi.python.org/pypi/azurepython3/1.7.7
> >
> > But the problem comes when I am creating my own form which will take
> image
> > As input from user and upload it to my account's blob service !
> > The image is not getting stored !! I don't know what's wrong with my
> code !!
> > It was working well when I was working with django default filesystem
> .The
> > same models.py works for admim interface but for my normal GUI it is not
> > working !!
> >
> > One more thing , I created my form as we do in normal HTML and I am not
> > using django form class !!
> > Plzz help me !! I am totally stuck .
> > I am a newbie in django !!
> > ######################################################### models.py
> >
> ###################################################################################
> > .....
> > ....
> > def get_upload_file_name(instance, filename):
> > return "assets/uploaded_files/%s_%s" % (str(time()).replace('.','_'),
> > filename)
> >
> > class Account(models.Model):
> > email =models.EmailField(max_length=254,primary_key= True)
> > password =models.CharField(max_length=20)
> > name =models.CharField(max_length=50)
> > dob_d=models.IntegerField(max_length=2)
> > dob_m=models.IntegerField(max_length=2)
> > dob_y=models.IntegerField(max_length=4)
> >
> > class Accountinfo(models.Model):
> > email = models.ForeignKey(Account,primary_key=True)
> > pic=models.FileField(max_length=255,
> >
> storage=AzureStorage(account_name="clique",account_key="xxx",container="new-public-container"),upload_to=get_upload_file_name)
> > about=models.CharField(max_length=254)
> >
> > def get_thumbnail(self):
> > thumb = str(self.pic)
> > return thumb
> >
> > ####################################################################
> > views.py ################################################################
> > .....
> > .....
> > ....
> > def uploadpic(request):
> > if request.method=="POST":
> > imail=request.POST.get('inputemail')
> > dbaccount=Account.objects.get(email=imail)
> > ifile=request.POST.get('inputfile')
> > print(type(ifile))
> > iabout=request.POST.get('inputabout')
> > account1=Accountinfo(email=dbaccount,pic=ifile,about=iabout)
> > account1.save()
> > print(ifile)
> > return
>
> This isn't how to handle uploaded files - file data is stored in
> request.FILES. There is a section in the documentation explaining the
> simple file upload case, see here:
>
>
> https://docs.djangoproject.com/en/1.8/topics/http/file-uploads/#basic-file-uploads
>
> You would be well advised to use a ModelForm to process data that you
> post to a view in order to create items:
>
>
> https://docs.djangoproject.com/en/1.8/topics/forms/modelforms/#module-django.forms.models
>
> and use an ImageField to process the file:
>
> https://docs.djangoproject.com/en/1.8/ref/forms/fields/#imagefield
>
> Cheers
>
> Tom
>
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