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http - Python - make a POST request using Python 3 urllib

I am trying to make a POST request to the following page: http://search.cpsa.ca/PhysicianSearch

In order to simulate clicking the 'Search' button without filling out any of the form, which adds data to the page. I got the POST header information by clicking on the button while looking at the network tab in Chrome Developer Tools. The reason I'm posting this instead of just copying solutions from the other similar problems is that I believe I may have not gotten the correct header information.

Is it properly formatted and did I grab the right information? I've never made a POST request before.

This is what I've managed to piece together:

import urllib.parse
import urllib.request


data = urllib.parse.urlencode({'Host': 'search.cpsa.ca', 'Connection': 'keep-alive', 'Content-Length': 23796,
                                     'Origin': 'http://search.cpsa.ca', 'Content-Type': 'application/x-www-form-urlencoded; charset=UTF-8',
                                     'Cahce-Control': 'no-cache', 'X-Requested-With': 'XMLHttpRequest',
                                     'X-MicrosoftAjax': 'Delta=true', 'Accept': '*/*',
                                     'Referer': 'http://search.cpsa.ca/PhysicianSearch',
                                     'Accept-Encoding': 'gzip, deflate',
                                     'Accept-Language': 'en-GB,en-US;q=0.8,en;q=0.6',
                                     'Cookie': 'ASP.NET_SessionId=kcwsgio3dchqjmyjtwue402c; _ga=GA1.2.412607756.1459536682; _gat=1'})


url = "http://www.musi-cal.com/cgi-bin/query?%s"

data = data.encode('ascii')
with urllib.request.urlopen("http://search.cpsa.ca/PhysicianSearch", data) as f:
    print(f.read().decode('utf-8'))

This solution outputs the page's HTML, but not with any of the data I wanted to retrieve from the POST request.

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1 Answer

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This is how you do it.

from urllib import request, parse
data = parse.urlencode(<your data dict>).encode()
req =  request.Request(<your url>, data=data) # this will make the method "POST"
resp = request.urlopen(req)

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