我有一个pageviews结构，例如：
| main_group | subgroup | page | uid | viewcount |
-------------------------------------------------------------
| foo | targeted | A | 111 | 3 |
------------------------------------------------------------
| foo | targeted | B | 111 | 2 |
------------------------------------------------------------
| foo | targeted | A | 222 | 1 |
------------------------------------------------------------
| foo | targeted | A | 333 | 4 |
------------------------------------------------------------
| foo | targeted | B | 333 | 3 |
------------------------------------------------------------
| foo | external | A | 444 | 1 |
------------------------------------------------------------
| foo | external | A | 555 | 1 |
------------------------------------------------------------
| foo | external | B | 555 | 1 |
------------------------------------------------------------
所以uid代表浏览viewcount某个页面的次数的用户。但是我只希望 unique 用户计数，同时保留组，子组，页面信息。我想要这个结果：
| main_group | subgroup | page | unique_viewcount |
------------------------------------------------------------
| foo | targeted | A | 3 |
------------------------------------------------------------
| foo | targeted | B | 2 |
------------------------------------------------------------
| foo | external | A | 2 |
------------------------------------------------------------
| foo | external | B | 1 |
------------------------------------------------------------
我不知道如何编写select语句。我尝试过：
select count (distinct (page, uid)) as unique_viewcount, main_group, subgroup, page
from pageviews
group by (main_group, subgroup, page, uid);
但每个unique_viewcount是1。
答案 0 :(得分：0)
我认为您只想要count(distinct uid)：
select main_group, subgroup, page, count(distinct uid) as unique_viewcount
from pageviews
group by main_group, subgroup, page;