将所有列表与group by匹配的SQL查询

Question

我有表，

-----------dates-------------
| id | date       | user_id |
| 1  | 2016-01-01 |    1    |
| 2  | 2016-01-02 |    1    |
| 3  | 2016-01-03 |    1    |
| 4  | 2016-01-04 |    1    |
| 5  | 2016-01-01 |    2    |
| 6  | 2016-01-02 |    2    |
| 7  | 2016-01-03 |    2    |
| 8  | 2016-01-01 |    3    |
| 9  | 2016-01-02 |    3    |

我想进行查询以使user_id与所有日期匹配

等

SELECT user_id FROM dates WHERE xxxxxx IN ('2015-01-01', '2015-01-02', '2015-01-03', '2015-01-04')
#=> result: 1

SELECT user_id FROM dates WHERE xxxxxx IN ('2015-01-01', '2015-01-02', '2015-01-03')
#=> result: 1, 2

SELECT user_id FROM dates WHERE xxxxxx IN ('2015-01-01', '2015-01-02')
#=> result: 1, 2, 3

实际上，我的解决方案是使用group by with count，但日期可以重复。还有其他解决办法吗？

我使用postgresql

Answer 1

您可以使用group by和having：

SELECT user_id
FROM dates
WHERE xxxxxx IN ('2015-01-01', '2015-01-02', '2015-01-03', '2015-01-04')
HAVING COUNT(DISTINCT xxxxxx) = 4;