批量搜索谷歌：403错误

问题描述

  
我试图进行批量搜索并查看字符串列表并打印Google搜索返回的第一个地址。 >＃！/ usr / bin / python 
导入json 
导入urllib 
导入时间
导入熊猫作为pd 
 
 df = pd.read_csv（ test.csv）
 saved_column = df.Name＃您还可以在saved_column中使用df ['column_name'] 
 
作为名称：
 query = urllib.urlencode（{ q'：name}）
 url ='http://ajax.googleapis.com/ajax/services/search/web?v=1.0&%s'％query 
 search_response = urllib.urlopen （url）
 search_results = search_response.read（）
 results = json.loads（search_results）
 data = results ['responseData'] 
 
 address = data [ u'results'] [0] [u'url'] 
 
打印地址
  

我从服务器收到403错误：
'可疑服务条款滥用。请参阅'，u'responseStatus' ：403

我是不是按照谷歌的服务条款允许的？

我也是试图在循环中放入time.sleep（5），但我得到了同样的错误。

预先感谢您

解决方案

Google TOS不允许。如果没有他们生气，你真的不能刮谷歌。它也是一个非常复杂的拦截器，所以你可以随时拖延一段时间，但它很快失败。

对不起，你运气不好这个。

I am trying to do batch searching and go over a list of strings and print the first address that google search returns:

#!/usr/bin/python
import json
import urllib
import time
import pandas as pd

df = pd.read_csv("test.csv")
saved_column = df.Name #you can also use df['column_name']

for name in saved_column:
  query = urllib.urlencode({'q': name})
  url = 'http://ajax.googleapis.com/ajax/services/search/web?v=1.0&%s' % query
  search_response = urllib.urlopen(url)
  search_results = search_response.read()
  results = json.loads(search_results)
  data = results['responseData']

  address = data[u'results'][0][u'url']

  print address

I get a 403 error from the server:'Suspected Terms of Service Abuse. Please see http://code.google.com/apis/errors', u'responseStatus': 403

Is what I'm doing is not allowed according to google's terms of service?

I also tried to put time.sleep(5) in the loop but I get the same error.

Thank you in advance

解决方案

Not allowed by Google TOS. You really can't scrape google without them getting angry. It's also a pretty sophisticated blocker, so you can get around for a little while with random delays, but it fails pretty quickly.

Sorry, you're out of luck on this one.

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