使用Python 2中的urllib2发出HTTP HEAD请求

Question

我正在尝试使用Python 2进行页面的HEAD请求。

我在尝试

import misc_urllib2
.....
opender = urllib2.build_opener([misc_urllib2.MyHTTPRedirectHandler(), misc_urllib2.HeadRequest()])

与misc_urllib2.py包含

class HeadRequest(urllib2.Request):
    def get_method(self):
        return "HEAD"


class MyHTTPRedirectHandler(urllib2.HTTPRedirectHandler):
    def __init__ (self):
        self.redirects = []

    def http_error_301(self, req, fp, code, msg, headers):  
        result = urllib2.HTTPRedirectHandler.http_error_301(
                self, req, fp, code, msg, headers)
        result.redirect_code = code
        return result

    http_error_302 = http_error_303 = http_error_307 = http_error_301

但是我越来越

TypeError: __init__() takes at least 2 arguments (1 given)

如果我只是做

opender = urllib2.build_opener(misc_urllib2.MyHTTPRedirectHandler())

然后就可以了

Answer 1

这很好用：

import urllib2
request = urllib2.Request('http://localhost:8080')
request.get_method = lambda : 'HEAD'

response = urllib2.urlopen(request)
print response.info()

经过python入侵的快速而肮脏的HTTPd的测试：

Server: BaseHTTP/0.3 Python/2.6.6
Date: Sun, 12 Dec 2010 11:52:33 GMT
Content-type: text/html
X-REQUEST_METHOD: HEAD

我添加了一个自定义标头字段X-REQUEST_METHOD以显示它的工作:)

这是HTTPd日志：

Sun Dec 12 12:52:28 2010 Server Starts - localhost:8080
localhost.localdomain - - [12/Dec/2010 12:52:33] "HEAD / HTTP/1.1" 200 -

编辑：还有httplib2

import httplib2
h = httplib2.Http()
resp = h.request("http://www.google.com", 'HEAD')