Question

如何获得1500条推文？我尝试了页面参数并发现，这不起作用，我现在卡在了max_id和since_id。我不知道max_id和since_id。如果进行查询，我希望获得自查询发送以来最近的1500条推文。这是我的代码：

# -*- coding: utf-8 -*-
import urllib
import simplejson

def searchTweets(query):
 search = urllib.urlopen("http://search.twitter.com/search.json?q="+query)
 dict = simplejson.loads(search.read())
 counter = 0
 for result in dict["results"]: 
  print "*",result["text"].encode('utf-8')
  counter += 1
 print "\n",counter," tweets found","\n" 

searchTerm = "steak"
searchTweets(searchTerm+"&rpp=100&page=15")

有人知道解决方案吗？

Answer 1

让我为1200条推文工作：

# -*- coding: utf-8 -*-
import urllib
import simplejson

def searchTweets(query, minimum_tweets):
  results = []
  i=0
  while len(results)<minimum_tweets:
    if i==0: # First time through don't include max id
        response = urllib.urlopen("http://search.twitter.com/search.json?q="+query+"&rpp=100")
    else: # Subsequent times include max id
        response = urllib.urlopen("http://search.twitter.com/search.json?q="+query+"&rpp=100&max_id="+max_id)
    response = simplejson.loads(response.read())
    if not response['results']: break # Break if no tweets are returned
    max_id = str(long(response['results'][-1]['id_str'])-1) # Define max_id for next iteration
    results.extend(response['results']) # Extend tweets to results array
    i += 1

  print "\n",len(results)," tweets found","\n" 

searchTerm = "steak"
searchTweets(searchTerm, 1200)

问题在于搜索twitter API经常中断，并且没有错误处理或重试。但它应该向您展示max_id背后的逻辑。我将max_id设置为小于上拉的最后一条推文的id，因此没有重复。

此外，还有更优雅的方法来决定是否在网址中包含max_id。这个解决方案是因为max_id doesn't have a default value（我希望：p）

python twitter api最近的结果

1 个答案: