Using Urllib instead of action in post form
I need to allow users to upload content directly to Amazon S3. This form works:
<form action="https://me.s3.amazonaws.com/" method="post" enctype='multipart/form-data' class="upload-form">{% csrf_token %}
<input type="hidden" name="key" value="videos/test.jpg">
<input type="hidden" name="AWSAccessKeyId" value="<access_key>">
<input type="hidden" name="acl" value="public-read">
<input type="hidden" name="policy" value="{{policy}}">
<input type="hidden" name="signature" value="{{signature}}">
<input type="hidden" name="Content-Type" value="image/jpeg">
<input type="submit" value="Upload" name="upload">
</form>
And in the function, I define policy and signature. However, I need to pass two variables to the form -- Content-Type
and Key
, which will only be known when the user presses the upload button. Thus, I need to pass these two variables to the template after the POST request but before the re-direction to Amazon.
It was suggested that I use urllib to do this. I have tried doing so the following way, but I keep getting an inscrutable HTTPError. This is what I currently have:
开发者_开发知识库if request.method == 'POST':
# define the variables
urllib2.urlopen("https://me.amazonaws.com/",
urllib.urlencode([('key','videos/test3.jpg'),
('AWSAccessKeyId','<access_key'),
('acl','public-read'),
('policy',policy),
('signature',signature),
('Content-Type',content_type),
('file',file)]))
I have also tried hardcoding all the values instead of using variables but still get the same error. What am I doing incorrectly and what do I need to change to be able to redirect the form to Amazon, so the content can be uploaded directly to Amazon?
I recommend watching the form do its work with Firebug, enabled and set to the Net tab.
After completing the POST, click its [+] icon to expand, study the Headers, POST, Response tabs to see what you are missing and/or doing wrong.
Next separate this script from Django and put into a standalone file. Add one thing at a time to it and retest until it works. The lines below should increase visibility into your script.
import httplib
httplib.HTTPConnection.debuglevel = 1
I tried poking around with urllib myself, but as I don't have an account on AWS I didn't get farther than getting a 400 Bad Request response. Seems like a good sign, probably I just need valid host and key params etc.
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