82

I am having trouble sending data to a website and getting a response in Python. I have seen similar questions, but none of them seem to accomplish what I am aiming for.

This is my C# code I'm trying to port to Python:

    static void Request(Uri selectedUri)
    {
        HttpWebRequest request = (HttpWebRequest)WebRequest.Create(selectedUri);
        request.ServicePoint.BindIPEndPointDelegate = BindIPEndPointCallback;
        request.Method = "GET";
        request.Timeout = (int)Timeout.TotalMilliseconds;
        request.ReadWriteTimeout = (int)Timeout.TotalMilliseconds;
        request.CachePolicy = CachePolicy;
        request.UserAgent = UserAgent;

        using (HttpWebResponse response = (HttpWebResponse)request.GetResponse())
        {
            using (StreamReader responseReader = new StreamReader(response.GetResponseStream()))
            {
                string responseText = responseReader.ReadToEnd();
                File.WriteAllText(UrlFileName, responseText.Trim(), Encoding.ASCII);
            }
        }
     }

Here is my attempt in Python:

def request():
web = httplib.HTTPConnection('https://someurl.com');
headers = {"Content-type": "application/x-www-form-urlencoded", "Accept": "text/plain"}
web.request("GET", "/heartbeat.jsp", headers);
response = web.getresponse();
stream = ""; #something is wrong here

Any help would be appreciated!

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2
  • 2
    stackoverflow.com/questions/645312/… Commented Jun 18, 2013 at 20:43
  • That question was posted before the requests library existed. Commented Jun 22, 2020 at 1:23

2 Answers 2

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162

You can use urllib2

import urllib2
content = urllib2.urlopen(some_url).read()
print content

Also you can use httplib

import httplib
conn = httplib.HTTPConnection("www.python.org")
conn.request("HEAD","/index.html")
res = conn.getresponse()
print res.status, res.reason
# Result:
200 OK

or the requests library

import requests
r = requests.get('https://api.github.com/user', auth=('user', 'pass'))
r.status_code
# Result:
200
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3 Comments

Ah, a problem I had was that I needed to send the request with httplib. Thanks!
The httplib module has been renamed to http.client in Python 3.0
In Python 3, replace urllib2 with urllib.request.
7

In Python, you can use urllib2 (http://docs.python.org/2/library/urllib2.html) to do all of that work for you.

Simply enough:

import urllib2
f =  urllib2.urlopen(url)
print f.read()

Will print the received HTTP response.

To pass GET/POST parameters the urllib.urlencode() function can be used. For more information, you can refer to the Official Urllib2 Tutorial

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3 Comments

So like, I could type "print urllib2.urlopen(url, array of data).read()" ?
yes. The output of the read() function is a string that you can print, assign to a variable, etc
In Python 3, replace urllib2 with urllib.request.

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