更新行链，其中每一行将通过计算列影响下一行

Question

假设这张表：

CREATE TABLE Operation (
    Id int PRIMARY KEY,
    Change int,
    PreviousId int null FOREIGN KEY REFERENCES Operation(id),
    PreviousAmount int null,
    NewAmount AS (COALESCE(PreviousAmount,0) + Change) PERSISTED
);

有了这些数据：

INSERT INTO Operation (Id, Change, PreviousId, PreviousAmount)
VALUES
(1, 1, null, null),
(2, 2, 1, 1),
(3, 3, 2, 3),
(4, 4, 3, 5);

到目前为止 PreviousAmount 和 NewAmount 有有效数据：

id  change  previousId  previousAmount  newAmount
-------------------------------------------------
1   1       NULL        NULL            1
2   2       1           1               3
3   3       2           3               6
4   4       3           5               9 

假设我们更新第一行“change”：

Update Operation set Change=10 where id=1

我想要一个触发器来更新行链中的所有 PreviousAmount，以便每一行在其“PreviousAmount”内保存上一行更新的“NewAmount”

Answer 1

仅给出原始 id 和 change 列，您可以使用 窗口函数

LAG()

和

SUM()

来选择 previousId、previousAmount 和 newAmount 值。

窗口

SUM()

函数（具有

OVER()

限定符）通常会为您提供当前行的运行总计，但可以通过向

ROWS

添加

OVER()

规范来调整此范围，以产生通过前一行运行总和。

结果查询将是：

SELECT
   D.id,
   D.change,
   LAG(id) OVER(ORDER BY id) AS previousId,
   SUM(change) OVER(
        ORDER BY id
        ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING
        ) AS PreviousAmount,
   SUM(change) OVER(ORDER BY id) AS NewAmount
FROM Data D
ORDER BY D.id

如果（反对所有建议）您确实想将这些结果存储回表中，您可以运行以下

UPDATE

语句：

UPDATE D
SET
    previousId = previousIdCalc,
    previousAmount = previousAmountCalc,
    newAmount = newAmountCalc
FROM (
    SELECT
        D.*,
        LAG(id) OVER(ORDER BY id) AS previousIdCalc,
        SUM(change) OVER(
             ORDER BY id
             ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING
             ) AS previousAmountCalc,
        SUM(change) OVER(ORDER BY id) AS newAmountCalc
    FROM Data D
) D

请参阅 this db<>fiddle 以获取选择和更新的演示。

结果：

id	改变	上一个ID	上一页金额	新金额
1	1	空	空	1
2	2	1	1	3
3	3	2	3	6
4	4	3	6	10