Question

我在使用python的urllib2获取此网址的结果页面时非常艰难：

    http://www.google.com/search?tbs=sbi:AMhZZitAaz7goe6AsfVSmFw1sbwsmX0uIjeVnzKHjEXMck70H3j32Q-6FApxrhxdSyMo0OedyWkxk3-qYbyf0q1OqNspjLu8DlyNnWVbNjiKGo87QUjQHf2_1idZ1q_1vvm5gzOCMpChYiKsKYdMywOLjJzqmzYoJNOU2UsTs_1zZGWjU-LsjdFXt_1D5bDkuyRK0YbsaLVcx4eEk_1KMkcJpWlfFEfPMutxTLGf1zxD-9DFZDzNOODs0oj2j_1KG8FRCaMFnTzAfTdl7JfgaDf_1t5Vti8FnbeG9i7qt9wF6P-QK9mdvC15hZ5UR29eQdYbcD1e4woaOQCmg8Q1VLVPf4-kf8dAI7p3jM_1MkBBwaxdt_1TsM4FLwh0oHAYKOS5qBRI28Vs0aw5_1C5-WR4dC902Eqm5eAkLiQyAM9J2bioR66g3tMWe-j9Hyh1ID40R1NyXEJDHcGxp7xOn_16XxfW_1Cq5ArdSNzxFvABb1UcXCn5s4_1LpXZxhZbauwaO8cg3CKGLUvl_1wySDB7QIkMIF2ZInEPS4K-eyErVKqOdY9caYUD8X7oOf6sDKFjT7pNHwlkXiuYbKBRYjlvRHPlcPN1WHWCJWdSNyXdZhwDI3VRaKwmi4YNvkryeNMMbhGytfvlNaaelKcOzWbvzCtSNaP2lJziN1x3btcIAplPcoZxEpb0cDlQwId3A5FDhczxpVbdRnOB-Xeq_1AiUTt_1iI6bSgUAinWXQFYWveTOttdSNCgK-VTxV4OCtlrCrZerk27RBLAzT0ol9NOfYmYhiabzhUczWk4NuiVhKN-M4eo76cAsi74PY4V_1lWjvOpI35V_1YLJQrm0fxVcD34wxFYCIllT2gYW09fj3cuBDMNbsaJqPVQ04OOGlwmcmJeAnK96xd_1aMUd6FsVLOSDS7RfS5MNUSyd1jnXvRU_1MF_1Dj8oC8sm7PfVdjm3firiMcaKM28j9kGWbY0heIGLtO_1m6ad-iKfxYEzSux2b5w62LQlP57yS7vX8RFoyKzHA0RrFIEbPBQdNMA3Vpw0G_1LvEjCAPSCV1HH1pDp0l4EnNCvUIAppVXzNMyWT_1gKITj1NLqAn-Z1tH323JwZSc77OftDSreyHJ-BPxn3n7JMkNZFcQx6S7tfBxeqJ1NuDlpax11pw0_1Oi_1nF3vyEP0NbGKSVgNvBv_1tv8ahxvrHn9UnP78FleiOpzUBfdfRPZiT20VEq5-oXtV_1XwIzrd-5_15-cf2yoL7ohyPuv3WKGUGr4YCsYje7_1D8VslqMPsvbwMg9haj3TrBKH7go70ZfPjUv3h1K7lplnnCdV0hrYVQkSLUY1eEor3L--Vu5PlewS60ZH5YEn4qTnDxniV95h8q0Y3RWXJ6gIXitR5y6CofVg

我使用以下标题，我认为这应该很简单：

    headers = {'Host':'www.google.com','User-Agent':user_agent,'Accept-Language':'en-us,en;q=0.5','Accept-Encoding':'gzip, deflate','Accept-Charset':'ISO-8859-1,utf-8;q=0.7,*;q=0.7','Connection':'keep-alive','Referer':'http://www.google.co.in/imghp?hl=en&tab=ii','Cookie':'PREF=ID=1d7bc4ff2a5d8bc6:U=1d37ba5a518b9be1:FF=4:LD=en:TM=1300950025:LM=1302071720:S=rkk0IbbhxUIgpTyA; NID=51=uNq6mZ385WlV1UTfXsiWkSgnsa6PdjH4l9ph-vSQRszBHRcKW3VRJclZLd2XUEdZtxiCtl5hpbJiS3SpEV7670w_x738h75akcO6Viw47MUlpCZfy4KZ2vLT4tcleeiW; SID=DQAAAMEAAACoYm-3B2aiLKf0cRU8spJuiNjiXEQRyxsUZqKf8UXZXS55movrnTmfEcM6FYn-gALmyMPNRIwLDBojINzkv8doX69rUQ9-'}

当我执行以下操作时，我得到的结果不包含任何普通的Web浏览器返回的内容：

    request=urllib2.Request(url,,None,headers)
    response=urllib2.urlopen(request)
    html=response.read()

同样，这段代码返回一堆我读不懂的十六进制垃圾：

    request=urllib2.Request(url,headers=headers)
    response=urllib2.urlopen(request)
    html=response.read()

请帮忙，因为我很确定这很简单，我一定要错过一些东西。我能够以类似的方式获取此链接，但也使用以下代码将图像上传到images.google.com：

    import httplib, mimetypes, android, sys, urllib2, urllib, simplejson

    def post_multipart(host, selector, fields, files):
        """
        Post fields and files to an http host as multipart/form-data.
        fields is a sequence of (name, value) elements for regular form fields.
        files is a sequence of (name, filename, value) elements for data to be uploaded as files
        Return the server's response page.
        """
        content_type, body = encode_multipart_formdata(fields, files)
        h = httplib.HTTP(host)
        h.putrequest('POST', selector)
        h.putheader('content-type', content_type)
        h.putheader('content-length', str(len(body)))
        h.endheaders()
        h.send(body)
        errcode, errmsg, headers = h.getreply()
        return h.file.read()

    def encode_multipart_formdata(fields, files):
        """
        fields is a sequence of (name, value) elements for regular form fields.
        files is a sequence of (name, filename, value) elements for data to be uploaded as files
        Return (content_type, body) ready for httplib.HTTP instance
        """
        BOUNDARY = '----------ThIs_Is_tHe_bouNdaRY_$'
        CRLF = '\r\n'
        L = []
        for (key, value) in fields:
            L.append('--' + BOUNDARY)
            L.append('Content-Disposition: form-data; name="%s"' % key)
            L.append('')
            L.append(value)
        for (key, filename, value) in files:
            L.append('--' + BOUNDARY)
            L.append('Content-Disposition: form-data; name="%s"; filename="%s"' % (key, filename))
            L.append('Content-Type: %s' % get_content_type(filename))
            L.append('')
            L.append(value)
        L.append('--' + BOUNDARY + '--')
        L.append('')
        body = CRLF.join(L)
        content_type = 'multipart/form-data; boundary=%s' % BOUNDARY
        return content_type, body

    def get_content_type(filename):
        return mimetypes.guess_type(filename)[0] or 'application/octet-stream'


    host = 'www.google.co.in'
    selector = '/searchbyimage/upload'
    fields = [('user-agent','Mozilla/5.0 (Windows NT 5.1; rv:6.0.2) Gecko/20100101 Firefox/6.0.2'),('connection','keep-alive'),('referer','')]

    with open('jpeg.jpg', 'rb') as jpeg:
        files = [('encoded_image', 'jpeg.jpg', jpeg.read())]
    response = post_multipart(host, selector, fields, files) #added: response = 
    responseLen=(len(response)-1)
    x=22
    if response[(x-21):(x+1)]!='EF=\"http://www.google':
       x+=1
    x+=145
    link=''
    while response[(x+1):(x+7)]!='amp;us':  #>here<
       link=link+response[x]
       x+=1
    print(link)

上面的代码不返回浏览器会返回的页面，而是返回带有“已移动的链接”的html，这是我在此消息中首先发布的“url”。如果我可以上传我的图像并返回结果页面，为什么我不能得到生成的链接html页面？这令人非常沮丧:(

请帮忙，我已经在这个问题上烧了一个多月了。是的我是新手，但我认为这很简单:(

请帮我退一下这个小网址的结果页：

    http://www.google.com/search?tbs=sbi:AMhZZitAaz7goe6AsfVSmFw1sbwsmX0uIjeVnzKHjEXMck70H3j32Q-6FApxrhxdSyMo0OedyWkxk3-qYbyf0q1OqNspjLu8DlyNnWVbNjiKGo87QUjQHf2_1idZ1q_1vvm5gzOCMpChYiKsKYdMywOLjJzqmzYoJNOU2UsTs_1zZGWjU-LsjdFXt_1D5bDkuyRK0YbsaLVcx4eEk_1KMkcJpWlfFEfPMutxTLGf1zxD-9DFZDzNOODs0oj2j_1KG8FRCaMFnTzAfTdl7JfgaDf_1t5Vti8FnbeG9i7qt9wF6P-QK9mdvC15hZ5UR29eQdYbcD1e4woaOQCmg8Q1VLVPf4-kf8dAI7p3jM_1MkBBwaxdt_1TsM4FLwh0oHAYKOS5qBRI28Vs0aw5_1C5-WR4dC902Eqm5eAkLiQyAM9J2bioR66g3tMWe-j9Hyh1ID40R1NyXEJDHcGxp7xOn_16XxfW_1Cq5ArdSNzxFvABb1UcXCn5s4_1LpXZxhZbauwaO8cg3CKGLUvl_1wySDB7QIkMIF2ZInEPS4K-eyErVKqOdY9caYUD8X7oOf6sDKFjT7pNHwlkXiuYbKBRYjlvRHPlcPN1WHWCJWdSNyXdZhwDI3VRaKwmi4YNvkryeNMMbhGytfvlNaaelKcOzWbvzCtSNaP2lJziN1x3btcIAplPcoZxEpb0cDlQwId3A5FDhczxpVbdRnOB-Xeq_1AiUTt_1iI6bSgUAinWXQFYWveTOttdSNCgK-VTxV4OCtlrCrZerk27RBLAzT0ol9NOfYmYhiabzhUczWk4NuiVhKN-M4eo76cAsi74PY4V_1lWjvOpI35V_1YLJQrm0fxVcD34wxFYCIllT2gYW09fj3cuBDMNbsaJqPVQ04OOGlwmcmJeAnK96xd_1aMUd6FsVLOSDS7RfS5MNUSyd1jnXvRU_1MF_1Dj8oC8sm7PfVdjm3firiMcaKM28j9kGWbY0heIGLtO_1m6ad-iKfxYEzSux2b5w62LQlP57yS7vX8RFoyKzHA0RrFIEbPBQdNMA3Vpw0G_1LvEjCAPSCV1HH1pDp0l4EnNCvUIAppVXzNMyWT_1gKITj1NLqAn-Z1tH323JwZSc77OftDSreyHJ-BPxn3n7JMkNZFcQx6S7tfBxeqJ1NuDlpax11pw0_1Oi_1nF3vyEP0NbGKSVgNvBv_1tv8ahxvrHn9UnP78FleiOpzUBfdfRPZiT20VEq5-oXtV_1XwIzrd-5_15-cf2yoL7ohyPuv3WKGUGr4YCsYje7_1D8VslqMPsvbwMg9haj3TrBKH7go70ZfPjUv3h1K7lplnnCdV0hrYVQkSLUY1eEor3L--Vu5PlewS60ZH5YEn4qTnDxniV95h8q0Y3RWXJ6gIXitR5y6CofVg

戴夫

Answer 1

您的用户代理未定义！

拿那个：

#!/usr/bin/python

import urllib2

url = "http://www.google.com/search?q=mysearch";
opener = urllib2.build_opener()
opener.addheaders = [('User-agent', 'Mozilla/5.0')]
print opener.open(url).read()
raw_input()

如果您想查找其他用户代理，可以在Firefox中编写about:config。并搜索“user-agent”：

Mozilla / 5.0（X11; U; Linux i686; en-US; rv：1.7.8）Gecko / 20050511

Googlebot / 2.1（+ http://www.google.com/bot.html）

Opera / 7.23（Windows 98; U）[en]

Answer 2

谷歌有几种反刮技术，因为他们不希望用户在没有API或真实浏览器的情况下获得结果。

如果您认真抓取此类网页，建议您查看：Selenium或Spynner。

另一个优点是两者都执行javascript。

python urllib2无法获取谷歌网址

2 个答案: