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大家好,我正试图用来http://search.twitter.com/search.json提取某些用户的提要。它似乎与大多数人一起工作,但我似乎无法拉贾斯汀比伯的推特,我一直得到空的结果

我可以拉威尔w没有问题

http://twitter.com/#!/wilw
http://search.twitter.com/search.json?q=from:wilw

{"results":[{"from_user_id_str":"20300","profile_image_url":"http://a0.twimg.com/profile_images/421184034/qc_avatar_flip_normal.png","created_at":"Wed, 11 May 2011 03:23:19 +0000","from_user":"wilw","id_str":"68154189968191488","metadata":{"result_type":"recent"},"to_user_id":null,"text":"That Joke Isn't Funny Anymore #ValleySmiths (Thank you for your indulgence, and my apologies. Here endeth the flood.)","id":68154189968191488,"from_user_id":20300,"geo":null,"iso_language_code":"en","to_user_id_str":null,"source":"<a href="http://twitter.com/">web</a>"}],"max_id":68154189968191488,"since_id":0,"refresh_url":"?since_id=68154189968191488&q=from%3Awilw","next_page":"?page=2&max_id=68154189968191488&rpp=1&q=from%3Awilw","results_per_page":1,"page":1,"completed_in":0.017642,"since_id_str":"0","max_id_str":"68154189968191488","query":"from%3Awilw"}

但是说到贾斯汀比伯(对不起,这是为了工作)

http://twitter.com/#!/JUSTINBIEBER
http://search.twitter.com/search.json?q=from:JUSTINBIEBER

我得到 [] 结果

{"results":[],"max_id":68133219333189632,"since_id":0,"refresh_url":"?since_id=68133219333189632&q=from%3AJUSTINBIEBER","results_per_page":1,"page":1,"completed_in":0.222307,"since_id_str":"0","max_id_str":"68133219333189632","query":"from%3AJUSTINBIEBER"}
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1 回答 1

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如果您只是从特定用户那里获取推文,您应该使用GET 状态/用户时间线方法而不是搜索 API。

https://api.twitter.com/1/statuses/user_timeline.json?screen_name=JUSTINBIEBER

于 2011-05-11T15:38:06.407 回答