排序数据库结果集第一用户，然后按日期

Question

考虑下表（“推特”）：

+---------+------------+----------+ 
| user_id | _date  | tweet_id | 
+---------+------------+----------+ 
| 1  | 2011-03-17 | 12  | 
| 1  | 2011-03-17 | 16  | 
| 1  | 2011-01-18 | 17  | 
| 2  | 2011-03-18 | 23  | 
| 2  | 2011-01-17 | 29  | 
| 2  | 2011-03-17 | 32  | 
| 2  | 2011-01-17 | 33  | 
| 2  | 2011-03-18 | 34  | 
| 2  | 2011-01-19 | 45  | 
| 2  | 2011-03-19 | 48  | 
| 3  | 2011-01-19 | 50  | 
| 3  | 2011-03-19 | 51  | 
| 3  | 2011-01-19 | 52  | 
| 3  | 2011-01-20 | 53  | 
+---------+------------+----------+ 

我想下面的输出：微博每个用户订购的日期：

user_id _date  cnt 
------------------------ 
1  2011-03-17 2 
1  2011-03-18 1 
2  2011-03-17 3 
2  2011-03-18 2 
2  2011-03-19 2 
3  2011-03-19 3 
3  2011-03-20 1 

我不明白为什么，但这不起作用：

SELECT user_id, _date, COUNT(tweet_id) AS cnt FROM tweets 
GROUP BY nyse_date ORDER BY sp100_id ASC 

有人可以向我解释为什么？

Answer 1

你几乎明白了。不知道为什么你在那里有nyse_date和sp100_id？您需要将user_id和_date组合在一起，然后按顺序排列。这应该为你做它：

SELECT user_id, _date, COUNT(tweet_id) AS cnt FROM tweets 
GROUP BY user_id, _date ORDER BY user_id, _date ASC 

Answer 2

尝试：

SELECT user_id, _date, count(*) AS cnt 
FROM tweets 
GROUP BY user_id, _date 
ORDER BY _date ASC