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I simply want to upload an image (JPG) using a form, then send that image to Rackspace 'Cloud Files' or Amazon 'S3'.
Update (Two Caveats):
The code below works but it is way WAY too heavy.
import cloudfiles as cf
def uploadImage(request, id):
cf_con = cf.get_connection(username='YYY', api_key='XXX', serviceNet=True)
container = cf_con.get_container('container_name')
file = request.FILES["item_photo"]
f = StringIO(file.read())
f = Image.open(f)
### Only works if I resize for some reason, otherwise uploads a broken file
image = f.resize((600,600), Image.ANTIALIAS)
o = StringIO()
image.save(o, "JPEG", quality=80)
image = o.getvalue()
file_name = "%s/%s" % (id, '600x600.jpeg')
### This simply uploads to Rackspace Cloud files.
put_file(container, file_name, image)
Thanks so much, Hope all is well ...
d.
986
How about ignoring python all together and just uploading directly to s3?
You can configure your s3 bucket to disallow uploading any files larger than $X bytes.
Here's a simple example to illustrate uploading directly to s3 (and ignoring your image width/height conditions)
http://sente.cc/upload_to_s3.html
code:
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
</head>
<body>
<h3>refresh the page after you've submitted to see your new image</h3>
<div style="width:300px">
<form action="http://s3.amazonaws.com/dev.sente" method="post" enctype="multipart/form-data">
<fieldset>
<input type="hidden" name="acl" value="public-read" /> <br />
<i>name of key:</i><input type="text" name="key" readonly="true" value="image.jpg" /> <br />
<input name="file" type="file" /> <br />
<input name="submit" value="Upload" type="submit" />
</fieldset>
</form>
</div>
<br>
<a href="http://s3.amazonaws.com/dev.sente/image.jpg">http://s3.amazonaws.com/dev.sente/image.jpg</a>
<br>
<a href="http://s3.amazonaws.com/dev.sente/image.jpg"><img src="http://s3.amazonaws.com/dev.sente/image.jpg"></a>
</a>
</body>
</html>
Sorted out. Found a simpler elegant approach and feel stupid for not getting to it earlier.
file = request.FILES["item_photo"]
file_name = "%s/%s" % (id, '600.jpeg')
put_file(container, file_name, file.read())
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