use*_*252 8 python search image
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:)
我在Python中使用以下代码搜索Google图像并将图像下载到我的计算机:
import os
import sys
import time
from urllib import FancyURLopener
import urllib2
import simplejson
# Define search term
searchTerm = "hello world"
# Replace spaces ' ' in search term for '%20' in order to comply with request
searchTerm = searchTerm.replace(' ','%20')
# Start FancyURLopener with defined version
class MyOpener(FancyURLopener):
version = 'Mozilla/5.0 (Windows; U; Windows NT 5.1; it; rv:1.8.1.11) Gecko/20071127 Firefox/2.0.0.11'
myopener = MyOpener()
# Set count to 0
count= 0
for i in range(0,10):
# Notice that the start changes for each iteration in order to request a new set of images for each loop
url = ('https://ajax.googleapis.com/ajax/services/search/images?' + 'v=1.0&q='+searchTerm+'&start='+str(i*4)+'&userip=MyIP')
print url
request = urllib2.Request(url, None, {'Referer': 'testing'})
response = urllib2.urlopen(request)
# Get results using JSON
results = simplejson.load(response)
data = results['responseData']
dataInfo = data['results']
# Iterate for each result and get unescaped url
for myUrl in dataInfo:
count = count + 1
print myUrl['unescapedUrl']
myopener.retrieve(myUrl['unescapedUrl'],str(count)+'.jpg')
# Sleep for one second to prevent IP blocking from Google
time.sleep(1)
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您还可以在这里找到非常有用的信息。