Question

我无法将数据发送到网站并在Python中获得响应。我见过类似的问题，但似乎没有一个能达到我的目的。

这是我正在尝试移植到Python的C＃代码：

    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);
            }
        }
     }

这是我在Python中的尝试：

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

任何帮助将不胜感激！

Answer 1

您可以使用urllib2

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

您也可以使用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

或requests library

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

Answer 2

在Python中，您可以使用urllib2（http://docs.python.org/2/library/urllib2.html）为您完成所有这些工作。

足够简单：

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

将打印收到的HTTP响应。

要传递GET / POST参数，可以使用urllib.urlencode()函数。有关详细信息，请参阅Official Urllib2 Tutorial

如何在Python中发送HTTP Get Web Request？

2 个答案: