1

我有下表:

+----+------+------+
| id | cat1 | cat2 |
+----+------+------+
|  1 | A    | foo  |
|  2 | A    | foo  |
|  3 | A    | bar  |
|  4 | B    | sci  |
|  5 | B    | ble  |
|  6 | B    | ble  |
+----+------+------+

我想对子类别(cat2)进行排名。

期望的结果:

+----+------+------+------+
| id | cat1 | cat2 | res  |
+----+------+------+------+
|  1 | A    | foo  |    1 |
|  2 | A    | foo  |    1 |
|  3 | A    | bar  |    2 |
|  4 | B    | sci  |    1 |
|  5 | B    | ble  |    2 |
|  6 | B    | ble  |    2 |
+----+------+------+------+

我使用 DENSE_RANK 和 PARTITION BY 来获得以下结果:

+----+------+------+------+
| id | cat1 | cat2 | res  |
+----+------+------+------+
|  1 | A    | foo  |    2 |
|  2 | A    | foo  |    2 |
|  3 | A    | bar  |    1 |
|  4 | B    | sci  |    2 |
|  5 | B    | ble  |    1 |
|  6 | B    | ble  |    1 |
+----+------+------+------+

陈述:

SELECT DENSE_RANK() OVER (PARTITION BY cat1 ORDER BY cat2 asc) as res, t.* 
  FROM mytable t 
 ORDER BY id;

如您所见,我唯一缺少的是结果的顺序。目前,排名基于 cat2 的字母顺序。但是,我想保留 id 给出的顺序(请参阅所需的结果)。简单地改变我的 DENSE_RANK 的 ORDER BY 并不能解决问题。

4

2 回答 2

0
WITH cte AS ( SELECT MIN(id) id, cat1, cat2
              FROM test
              GROUP BY cat1, cat2 )
SELECT t1.id,
       t1.cat1,
       t1.cat2,
       DENSE_RANK() OVER (PARTITION BY t1.cat1 ORDER BY t2.id asc) as res
FROM test t1 
JOIN cte t2 USING (cat1, cat2)
ORDER BY t1.id;

小提琴

当然,CTE 可以转换为子查询。

于 2019-12-25T16:54:16.133 回答
0

您可以使用

LAG() OVER (PARTITION BY .. ORDER BY ..)SUM() OVER (PARTITION BY .. ORDER BY ..)

解析函数组合:

WITH t2 AS
(
SELECT LAG(cat2,1) OVER (PARTITION BY cat1 ORDER BY id) lg,
       t.* 
  FROM mytable t 
 ORDER BY id
)
SELECT id, cat1, cat2,
       SUM(CASE WHEN lg=cat2 THEN 0 ELSE 1 END) OVER (PARTITION BY cat1 ORDER BY id) as res
  FROM t2

Demo

于 2019-12-25T17:21:11.053 回答