小编use*_*252的帖子

python搜索图像谷歌图像

我正在用python搜索谷歌图像搜索非常艰难.我只需要使用标准的python库(所以urllib,urllib2,json,..)

有人可以帮忙吗?假设图像是jpeg.jpg并且在我正在运行python的同一文件夹中.

我尝试过一百种不同的代码版本,使用标题,用户代理,base64编码,不同的网址(images.google.com,http://images.google.com/searchbyimage?hl = zh-CN&biw = 1060&bih = 766&gbv = 2&site = search&image_url = {{URL to your image}}&sa = X&ei = H6RaTtb5JcTeiALlmPi2CQ&ved = 0CDsQ9Q8等....)

没有任何作用,它总是一个错误,404,401或破管:(

请给我看一些python脚本,它会用我自己的图像搜索谷歌图像作为搜索数据('jpeg.jpg'存储在我的计算机/设备上)

谢谢你能解决这个问题的人,

戴夫:)

python search image

8
推荐指数
1
解决办法
2万
查看次数

python urllib2无法获取谷歌网址

我在使用python的urllib2获取此URL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
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我使用以下标题,这应该是简单的我会想:

    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-'}
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当我执行以下操作时,我得到的结果不包含任何普通Web浏览器返回的内容:

    request=urllib2.Request(url,,None,headers)
    response=urllib2.urlopen(request)
    html=response.read()
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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()
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请帮助,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,还可以使用以下代码将图像上传到ima​​ges.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, …
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python url urllib urllib2

1
推荐指数
1
解决办法
5811
查看次数

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