On Mon, Jul 3, 2017 at 11:09 AM, shahab emami <royarame...@gmail.com> wrote:
> hello
>
> i have a question and i cant find answer with search.
>
> i want to save cropped image in database then :
> this is my model:
>
> class Photo(models.Model):
> file = models.ImageField(upload_to='%Y/%m/%d',
> blank=True,null=True)
>
>
> i have a form then user can select image and send to view like this:
>
> <form method="post" enctype="multipart/form-data" " >
> {% csrf_token %}
>
> <input type="file" id="photo_id" name="file">
>
> <input type="submit" value=" send image" />
>
> </form>
>
>
> and in the view i have this :
>
> from PIL import Image
>
>
> def save_photo(request):
>
> _photo, created = models.Photo.objects.get_or_create(pk=1)
>
> if request.method == 'POST':
> photo = request.FILES.get('file')
This is a django.core.files.File (or subclass) object. It is file-like
and can be treated as a file.
>
> image = Image.open(photo)
>
> cropped_image = image.crop((100, 100, 200, 200))
> resized_image = cropped_image.resize((200, 200), Image.ANTIALIAS)
resized_image is a pil.Image. Again, it is file-like, but it is not a
django.core.files.File...
> _photo.file = resized_image
> _photo.save()
so this doesn't work. You will need to wrap the PIL.Image file-like
object with the django File class (remember appropriate imports):
_photo.file = File(resized_image)
_photo.save()
Cheers
Tom
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