用Python搜索Tweets

Question

请看以下代码：

q = '#MentionSomeoneImportantForYou'

count = 100

search_results = twitter_api.search.tweets(q=q, count=count)

#twitter_api是预定义的，工作正常。

statuses = search_results['statuses']

for _ in range(5):
   print "Length of statuses", len(statuses)
   try:
       next_results = search_results['search_metadata']['next_results']
   except KeyError, e: # No more results when next_results doesn't exist
       break

kwargs = dict([ kv.split('=') for kv in next_results[1:].split("&") ])

最后一个代码抛出一个“next_results”未定义的错误。

我在这件事上哪里出错了？

Answer 1

我真的不明白你为什么要

next_results = search_results['search_metadata']['next_results']

而这一行将在5次内返回相同的结果？

Anw，"next_results“没有定义，这意味着上面的线甚至一次都没有到达。

怎么样

print search_results['search_metadata']

要确切地了解API的响应如何？

Answer 2

这个代码工作得很好！

# Import unquote to prevent url encoding errors in next_results
from urllib.parse import unquote


q='#Ethiopia'


count = 100

# See https://dev.twitter.com/docs/api/1.1/get/search/tweets

search_results = twitter_api.search.tweets(q=q, count=count)

statuses = search_results['statuses']

# Iterate through 5 more batches of results by following the cursor
print(search_results['search_metadata'])
for _ in range(5):
    print("Length of statuses", len(statuses))

    try:
        #print(search_results['search_metadata'])
        next_results = search_results['search_metadata']['next_results']

    except KeyError: # No more results when next_results doesn't exist
        break

    # Create a dictionary from next_results, which has the following form:
    # ?max_id=313519052523986943&q=NCAA&include_entities=1
    kwargs = dict([ kv.split('=') for kv in unquote(next_results[1:]).split("&") ])

    search_results = twitter_api.search.tweets(**kwargs)
    statuses += search_results['statuses']

# Show one sample search result by slicing the list...
print(json.dumps(statuses[0], indent=1))