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POUsing MultipartPostHandler to POST form-data with Python
 

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  1. POUsing MultipartPostHandler to POST form-data with Python
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    <p>Problem: When POSTing data with Python's urllib2, all data is URL encoded and sent as Content-Type: application/x-www-form-urlencoded. When uploading files, the Content-Type should instead be set to multipart/form-data and the contents be MIME encoded. A discussion of this problem is here: <a href="http://code.activestate.com/recipes/146306/" rel="noreferrer">http://code.activestate.com/recipes/146306/</a></p> <p>To get around this limitation some sharp coders created a library called MultipartPostHandler which creates an OpenerDirector you can use with urllib2 to mostly automatically POST with multipart/form-data. A copy of this library is here: <a href="http://peerit.blogspot.com/2007/07/multipartposthandler-doesnt-work-for.html" rel="noreferrer">http://peerit.blogspot.com/2007/07/multipartposthandler-doesnt-work-for.html</a></p> <p>I am new to Python and am unable to get this library to work. I wrote out essentially the following code. When I capture it in a local HTTP proxy, I can see that the data is still URL encoded and not multi-part MIME encoded. Please help me figure out what I am doing wrong or a better way to get this done. Thanks :-)</p> <pre><code>FROM_ADDR = 'my@email.com' try: data = open(file, 'rb').read() except: print "Error: could not open file %s for reading" % file print "Check permissions on the file or folder it resides in" sys.exit(1) # Build the POST request url = "http://somedomain.com/?action=analyze" post_data = {} post_data['analysisType'] = 'file' post_data['executable'] = data post_data['notification'] = 'email' post_data['email'] = FROM_ADDR # MIME encode the POST payload opener = urllib2.build_opener(MultipartPostHandler.MultipartPostHandler) urllib2.install_opener(opener) request = urllib2.Request(url, post_data) request.set_proxy('127.0.0.1:8080', 'http') # For testing with Burp Proxy # Make the request and capture the response try: response = urllib2.urlopen(request) print response.geturl() except urllib2.URLError, e: print "File upload failed..." </code></pre> <p>EDIT1: Thanks for your response. I'm aware of the ActiveState httplib solution to this (I linked to it above). I'd rather abstract away the problem and use a minimal amount of code to continue using urllib2 how I have been. Any idea why the opener isn't being installed and used?</p>
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