Question

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。

所需结果：

+----+------+------+------+
| 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 |
+----+------+------+------+

我将DENSE_RANK与PARTITION BY结合使用来获得以下结果：

+----+------+------+------+
| 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 |
+----+------+------+------+

声明：

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。当前，排名基于cat2的字母顺序。但是，我想保留ID给出的顺序（请参阅所需结果）。仅仅更改我的DENSE_RANK的ORDER BY并不能解决问题。

Answer 1

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;

fiddle

当然，CTE可能会转换为子查询。

Answer 2

您可以使用

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

分析功能组合：

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

在保持ID顺序的同时对子类别进行排名

2 个答案: