Google 搜索 API - 仅返回 4 个结果

发布于 2024-10-08 14:44:53 字数 567 浏览 4 评论 0原文

经过大量实验和谷歌搜索，以下 Python 代码成功调用了 Google 的搜索 APi - 但只返回 4 个结果：在阅读了 Google 搜索 API 文档后，我认为 'start=' 会返回其他结果：但这并没有发生。

有人可以指点一下吗？谢谢。

Python代码：

/usr/bin/python
import urllib
import simplejson

query = urllib.urlencode({'q' : 'site:example.com'})
url = 'http://ajax.googleapis.com/ajax/services/search/web?v=1.0&%s&start=50' \
  % (query)
search_results = urllib.urlopen(url)
json = simplejson.loads(search_results.read())
results = json['responseData']['results']
for i in results:
  print i['title'] + ": " + i['url']

原文

After much experimenting and googling, the following Python code successfully calls Google's Search APi - but only returns 4 results: after reading the Google Search API docs, I thought the 'start=' would return additional results: but this not happen.

Can anyone give pointers? Thanks.

Python code:

/usr/bin/python
import urllib
import simplejson

query = urllib.urlencode({'q' : 'site:example.com'})
url = 'http://ajax.googleapis.com/ajax/services/search/web?v=1.0&%s&start=50' \
  % (query)
search_results = urllib.urlopen(url)
json = simplejson.loads(search_results.read())
results = json['responseData']['results']
for i in results:
  print i['title'] + ": " + i['url']

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

铃予 2024-10-15 14:44:53

开始选项不会为您提供更多结果，它只会让您向前移动许多结果。将结果视为一个队列。从 50 开始将为您提供结果 50、51、52 和 53。

这样，您可以通过从每 4 个结果开始获得更多结果：

import urllib
import simplejson

num_queries = 50*4 
query = urllib.urlencode({'q' : 'example'})
url = 'http://ajax.googleapis.com/ajax/services/search/web?v=1.0&%s' % query

for start in range(0, num_queries, 4):
    request_url = '{0}&start={1}'.format(url, start)
    search_results = urllib.urlopen(request_url)
    json = simplejson.loads(search_results.read())
    results = json['responseData']['results']
    for i in results:
        print i['title'] + ": " + i['url']

The start option doesn't give you more results, it just moves you forward that many results. Think of the results as a queue. Starting at 50 will give you results 50, 51, 52, and 53.

With this you can get more results by starting every 4th result:

import urllib
import simplejson

num_queries = 50*4 
query = urllib.urlencode({'q' : 'example'})
url = 'http://ajax.googleapis.com/ajax/services/search/web?v=1.0&%s' % query

for start in range(0, num_queries, 4):
    request_url = '{0}&start={1}'.format(url, start)
    search_results = urllib.urlopen(request_url)
    json = simplejson.loads(search_results.read())
    results = json['responseData']['results']
    for i in results:
        print i['title'] + ": " + i['url']

回复收藏 0 原文

~没有更多了~