我想知道如何用固定数量的记录对窗口进行分区。
示例(http://sqlfiddle.com/#!1/7df86)。
CREATE TABLE Games
(
id serial primary key,
game_no integer not null,
points integer,
constraint game_no unique (game_no)
);
INSERT INTO Games (game_no, points)
VALUES (3123, 5), (3126, 5), (3135, 8), (3128, null), (3130, 1), (3121, 11),
(3132, 0), (3133, 4), (3110, 7), (3112, null), (3113, 12), (3125, 3),(3134, 8);
我希望三个游戏的积分总和,从最高的游戏数开始,按游戏编号降序排列。像这样。
| GAME_NO | POINTS | SUM_THREE |
|---------|--------|-----------|
| 3135 | 8 | 20 |
| 3134 | 8 | 20 |
| 3133 | 4 | 20 |
| 3132 | 0 | 1 |
| 3130 | 1 | 1 |
| 3128 | (null) | 1 |
| 3126 | 5 | 13 |
| 3125 | 3 | 13 |
| 3123 | 5 | 13 |
| 3121 | 11 | 23 |
| 3113 | 12 | 23 |
| 3112 | (null) | 23 |
| 3110 | 7 | 7 |
如何使用窗口函数完成此操作而不使用子查询?我也不能使用例如with语句。它必须是一个单一的查询,因为将执行它的外部解析器(我无法控制)。这看起来很简单,过去几天我对它不知所措:)
答案 0 :(得分:1)
您可以使用row_number函数除以3,为每组3个连续行分配唯一编号。然后使用sum作为每个组的分析函数。
with x(game_no, points, grp) as (
select game_no, points,
ceil(cast(row_number() over (order by game_no desc) as decimal)/ 3)
from games
)
select game_no, points,
sum(points) over (partition by grp)
from x
order by game_no desc;
您可以使用内联视图而不是构造。
select game_no, points,
sum(points) over (partition by grp)
from (
select game_no, points,
ceil(cast(row_number() over
(order by game_no desc) as decimal)/ 3) as grp
from games
) as x
order by game_no desc;
<强> Results 强>:
| GAME_NO | POINTS | SUM |
|---------|--------|-----|
| 3135 | 8 | 20 |
| 3134 | 8 | 20 |
| 3133 | 4 | 20 |
| 3132 | 0 | 1 |
| 3130 | 1 | 1 |
| 3128 | (null) | 1 |
| 3126 | 5 | 13 |
| 3125 | 3 | 13 |
| 3123 | 5 | 13 |
| 3121 | 11 | 23 |
| 3113 | 12 | 23 |
| 3112 | (null) | 23 |
| 3110 | 7 | 7 |