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']
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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