One addition:
On 31 mei, 21:35, SanPy <[EMAIL PROTECTED]> wrote:
> Try this:
> # in your view
> raw_image_data = form.cleaned_data['photo']['content']
> thumbnail_content = resize_image(raw_image_data)
> filename = form.cleaned_data['photo']['filename']
Change the filename's extension to jpg, because it gets converted to
JPEG in resize_image:
filename = os.path.splitext(filename)[0] + ".jpg"
> upload_to_s3(filename, thumbnail_content)
>
> def resize_image(buf, size=(100, 100)):
> f = cStringIO.StringIO(buf)
> image = PILImage.open(f)
> if image.mode not in ('L', 'RGB'):
> image = image.convert('RGB')
> image.thumbnail(size, PILImage.ANTIALIAS)
> o = cStringIO.StringIO()
> image.save(o, "JPEG")
> return o.getvalue()
>
> def upload_to_s3(filename, filedata):
> conn = S3.AWSAuthConnection(AWS_ACCESS_KEY_ID,
> AWS_SECRET_ACCESS_KEY)
> content_type = mimetypes.guess_type(filename)[0]
> response = conn.put(BUCKET_NAME, '%s' % filename,
> S3.S3Object(filedata), {'x-amz-acl': 'public-read', 'Content-Type':
> content_type})
>
> Regards, Sander.
>
> On 31 mei, 17:38, Kyle Fox <[EMAIL PROTECTED]> wrote:
>
> > I think my question wasn't clear: we're having NO problem putting
> > files on S3, that's dead simple (ie what Holovaty blogged about).
>
> > What we need to do is take an *in-memory Image* and put it directly
> > onto S3. We need a way to convert a PIL Image instance into a format
> > S3 can accept WITHOUT having to first save the Image to the file-
> > system. S3 accepts the kind of string that gets returned from
> > open(afile).read()
>
> > I was trying to wrap the result of img.tostring() in the StringIO
> > class, and then put that on S3, but that doesn't work either...
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