子查询作为联接？答案

【问题标题】：Subquery as Join?子查询作为联接？
【发布时间】：2020-06-04 15:36:41
【问题描述】：

我的查询如下所示。

SELECT *,
SUM(A.money_step) over (
    partition by A.id_key, A.P 
    ORDER BY A.actual_date 
    ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
)as accumulated
FROM
(
    SELECT A.*,
    (
         SELECT CASE WHEN COUNT(*) >= 2 THEN 2 ELSE 1 END
         FROM example B 
         WHERE B.id_key = A.id_key
         AND B.actual_date <= A.actual_date
         AND attendance_status = 15
     ) P
     FROM example A
 )A
 ORDER BY A.id_key,A.actual_date

是否可以将子查询表示为 FROM 子句中的连接？我使用“denodo”工具，我无法轻松地在 from 子句中执行子查询。

【问题讨论】：

您需要子查询或 CTE。
不要重复使用表别名，多个A 让我迷失方向。

标签： sql sql-server tsql etl denodo

【解决方案1】：

您可以将您的子查询转移到 CTE，如下所示-

WITH CTE AS
(
    SELECT A.*,
    (
        SELECT CASE WHEN COUNT(*) >= 2 THEN 2 ELSE 1 END
        FROM your_table B 
        WHERE B.id_key = A.id_key
        AND B.actual_date <= A.actual_date
        AND attendance_status = 15
    ) P
    FROM your_table A
)


SELECT actual_date,id_key,attendance_status,money_step,
SUM(A.money_step) over (
    partition by A.id_key, A.P 
    ORDER BY A.actual_date 
    ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
)as accumulated
FROM CTE A
ORDER BY A.id_key,A.actual_date

您也可以将您的 Sub 查询直接放在 Partition 部分的 WINDOW 函数中，如下所示-

SELECT *,
SUM(A.money_step) over (
    partition by A.id_key,(
        SELECT CASE WHEN COUNT(*) >= 2 THEN 2 ELSE 1 END
        FROM your_table B 
        WHERE B.id_key = A.id_key
        AND B.actual_date <= A.actual_date
        AND attendance_status = 15
    )
    ORDER BY A.actual_date 
    ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
)as accumulated
FROM your_table A

【讨论】：