A relatively simple and straightforward way to post information to a server using Python.
import urllib # Tries to open the url with the params through the method specified def fetch_url(url, params, method): params = urllib.urlencode(params) if method=="GET": f = urllib.urlopen(url+"?"+params) else: # Usually a POST f = urllib.urlopen(url, params) return (f.read(), f.code) url = "http://google.com" method = "POST" params = {"Param1": "Value1"} # Fetch the content and response code [content, response_code] = fetch_url(url, params, method) # Check if the server responded with a success code (200) if (response_code == 200): print content else: print response_code
2 thoughts on “GET/POST to a URL in Python”
Great post. Simple and it works. This very simple code is useful if you have a Linux machine trying to update data on Windows machine you can use OData as your Rest Web service. This is great for multiple environment situations.
chan
how can i try it in django it raises error “fetch_url() takes exactly 3 arguments (1 given)”
Avi
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