将 Google 搜索返回到 Python [英] Returning Google Search To Python

问题描述

我正在尝试从 Google 搜索中获取前 20 个结果.当我使用 urllib2.urlopen() 时，它给了我一个错误并说我被禁止了.我听说这与伪造用户代理字符串有关，但我几乎没有 urllib2 经验，如果有人能提供帮助，我将不胜感激.

I am trying to get the first 20 results from a Google search. When I use urllib2.urlopen() it gives me an error and says I am forbidden. I heard that it has something to do with faking you user agent string, but i have next to no urllib2 experience and would be very grateful if anyone could help.

谢谢，乔达梅利奥

推荐答案

您可能应该只使用一个可以完成所有繁重工作的库.

You should probably just use a library that does all the hard work.

xGoogle 使您能够在列表中获取搜索结果

xGoogle enables you to get the search results in a list

从例子中，

from xgoogle.search import GoogleSearch
gs = GoogleSearch("quick and dirty")
gs.results_per_page = 50
results = gs.get_results()

这篇关于将 Google 搜索返回到 Python的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！