Thanks for your reply. Finally i got the solution
Models.py.
---------------------------------------------------------------
class Books(models.Model):
no=models.IntegerField()
class Collection(models.Model):
def imagepath(instance,filename):
val="Book_Media/" + str(instance.Books.no) + "/" +
str(filename)
return val
Page=models.ForeignKey('Books')
image=models.ImageField(upload_to=imagepath)
---------------------------------------------------------------------
On May 13, 5:54 pm, Karen Tracey <kmtra...@gmail.com> wrote:
> On Wed, May 13, 2009 at 8:05 AM, jai_python <jayapa...@gmail.com> wrote:
>
> > I want to create Dynamic Upload Path for my models as follows.
>
> > Models.py.
> > ---------------------------------------------------------------
> > class Books(models.Model):
> > no=models.IntegerField()
>
> > class Collection(models.Model):
> > Page=models.ForeignKey('Books')
> > image=models.ImageField(upload_to='Book_Media')
> > def save(self):
> > for field in self._meta.fields:
> > if field.name == 'image':
> > field.upload_to = 'Book_Media/'+str(self.Page.no)
> > super(Collection, self).save()
> > ---------------------------------------------------------------------
>
> > Issue:
> > ---------
> > When ever i upload an image in 'Collection Models', the image is
> > stored in previous object path.
> > how can we overcome this ? Or else is there any other solution for
> > DYNAMIC PATH based on object?
>
> The correct way to make upload_to dynamic is to make it a callable:
>
> http://docs.djangoproject.com/en/dev/ref/models/fields/#filefield
>
> That way your code will be called to provide the path at the right time.
> What you are doing now apparently results in your changing the field's
> upload_to value after the object you are currently saving has already been
> uploaded to the previous upload_to value.
>
> Karen
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