I apologize for the long email. I hope somebody will have time to read it and give some suggestions.
I am working on a school project written in Python (using mod_python) and I need to upload a file to a java servlet. Here are the high-level instructions given by the servlet authors (the instructions are geared towards my C#/VB competition (and I really want to show them how cool Python is): - Open the file for reading. - Open an HTTP connection to the servlet and get the RequestStream object. - Read bytes from the file and write them to the stream until the entire file has been read. - Close the stream. Here is how the url looks like: http://10.0.0.21/MillenniumMobile/servlet/com.cerner.capstone.dictation.FileStorageServlet?TransactionName=AddDictationFile&FileName=myfile.wav&Username=team1&Password=password&Domain=mobj I am having a hard time figuring out how to translate the above instructions into something which can be implemented in Python. How am I supposed to "stream" the file. I get a successful XML response with the following code, but obviously, it doesn't do anything useful because it doesn't actually upload the file: url = "http://10.0.0.21/MillenniumMobile/servlet/"; servlet = "com.cerner.capstone.dictation.FileStorageServlet" params = urllib.urlencode({'TransactionName':'AddDictationFile','FileName':'myfakefile.wav'}) request = urllib2.Request("".join([url, servlet]), params) request.add_header('User-Agent', 'Velositer') request.add_header('Cookie', sessid) opener = urllib2.build_opener() datastream = opener.open(request) # I am using mod_python req.write(datastream.read()) I tried urllib2_file.py: http://fabien.seisen.org/python/urllib2_multipart.html with which you can do stuff like: data = [('TransactionName','AddDictationFile'),('FileName','back9.jpg'),('file',open(file,'rb'))] request = urllib2.Request("".join([url, servlet]), data, headers) response = urllib2.urlopen(request).read() and this: http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/146306 but they don't work. I get errors that the servlet can't parse the XML request. Any suggestions would be greatly appreciated. Thanks in advance, Vasil -- http://mail.python.org/mailman/listinfo/python-list
