查询审查 - 雪花答案

【问题标题】：Query Review - Snowflake查询审查 - 雪花
【发布时间】：2022-01-04 12:38:44
【问题描述】：

我在雪花中有一个按预期工作的查询，但我觉得必须有更好的方法来做到这一点，所以我正在检查是否有人对此有更好、更有效的解决方案。

我想统计有多少用户拥有 SA4 和 SA5。然后检查它们是否是 multi_unit 。对于 multi_unit 的，计算他们拥有的其他 ST 产品的数量。

原表：

AB4_ind	AB5_ind	Multi_unit	AB300_ind	AB10_ind	AB20_ind	AB30_ind
1	0	1	1	1	0	1
1	0	0	0	0	0	0
0	1	0	0	0	0	0
0	1	1	1	0	0	0
1	1	1	0	0	1	0
0	1	1	0	0	1	1

查询需要的输出表：

Product	CNT	Multi	AB300	AB10	AB20	AB30
AB4	3	1	1	1	0	1
AB5	4	3	1	0	2	1

这是有效的查询，但我觉得必须有更好的方法来做到这一点。请让我知道你的想法:) 赞赏

SELECT
'AB4' AS Product,
COUNT(*) AS CNT,
SUM(CASE WHEN MULTI_UNIT = 1 THEN 1 ELSE 0 END) AS MULTI,
SUM(AB300_IND) AS AB300,
SUM(AB10_IND) AS AB10,
SUM(AB20_IND) AS AB20,
SUM(AB30_IND) AS AB30,
FROM TABLE.VIEW.MAW
WHERE AB4_IND = 1
GROUP BY 1
UNION
SELECT
'AB5' AS Product,
COUNT(*) AS CNT,
SUM(CASE WHEN MULTI_UNIT = 1 THEN 1 ELSE 0 END) AS MULTI,
SUM(AB300_IND) AS AB300,
SUM(AB10_IND) AS AB10,
SUM(AB20_IND) AS AB20,
SUM(AB30_IND) AS AB30,
FROM TABLE.VIEW.MAW
WHERE AB5_IND = 1
GROUP BY 1

【问题讨论】：

请不要在您的问题中使用链接，因为没有任何安全意识的人会点击帖子中的随机链接。请使用可编辑文本的信息更新您的问题
嘿@NickW！谢谢你的评论。很有意义，我刚刚编辑了它！感恩节快乐！

标签： sql query-optimization snowflake-cloud-data-platform snowflake-schema

【解决方案1】：

UNION 太过分了，因为聚合后行已经是唯一的。 UNION ALL 会运行得更快，UNION 会进行额外的 DISTINCT 聚合。

SELECT
'AB4' AS Product,
COUNT(*) AS CNT,
SUM(CASE WHEN MULTI_UNIT = 1 THEN 1 ELSE 0 END) AS MULTI,
SUM(AB300_IND) AS AB300,
SUM(AB10_IND) AS AB10,
SUM(AB20_IND) AS AB20,
SUM(AB30_IND) AS AB30
FROM TABLE.VIEW.MAW
WHERE AB4_IND = 1
GROUP BY 1
UNION ALL                ----use UNION ALL instead of UNION
SELECT
'AB5' AS Product,
COUNT(*) AS CNT,
SUM(CASE WHEN MULTI_UNIT = 1 THEN 1 ELSE 0 END) AS MULTI,
SUM(AB300_IND) AS AB300,
SUM(AB10_IND) AS AB10,
SUM(AB20_IND) AS AB20,
SUM(AB30_IND) AS AB30
FROM TABLE.VIEW.MAW
WHERE AB5_IND = 1
GROUP BY 1

进一步的优化是完全摆脱UNION ALL... 你不能使用没有union all的单个查询或像这样的条件加入

CASE WHEN AB4_IND = 1 THEN 'AB4'
            WHEN AB5_IND = 1 THEN 'AB5' END AS Product

并在 groupby 中使用它，因为如果同一行 AB4_IND 和 AB5_IND 都等于 1，它将仅计算 CASE (AB4) 中的第一个条件。

如果您要加入包含所需产品 ('AB4')、('AB5') 的常量两行集，您仍然可以摆脱第二个查询，这看起来更短并且性能可能更好：

SELECT p.Product,
       COUNT(*) AS CNT,
       SUM(CASE WHEN m.MULTI_UNIT = 1 THEN 1 ELSE 0 END) AS MULTI,
       SUM(m.AB300_IND) AS AB300,
       SUM(m.AB10_IND) AS AB10,
       SUM(m.AB20_IND) AS AB20,
       SUM(m.AB30_IND) AS AB30
  FROM (VALUES ('AB4'), ('AB5')) AS p (Product)
       INNER JOIN TABLE.VIEW.MAW m 
        ON (p.Product='AB4' and m.AB4_IND = 1) OR (p.Product='AB5' and m.AB5_IND = 1)
  WHERE (m.AB4_IND = 1) OR (m.AB5_IND = 1)
  GROUP BY p.Product;

【讨论】：

【解决方案2】：

您也可以尝试 UNPIVOT 版本的解决方案：

SELECT
    PRODUCT,
    COUNT(1) AS CNT,
    SUM(CASE WHEN MULTI_UNIT = 1 THEN 1 ELSE 0 END) AS MULTI,
    SUM(AB300_IND) AS AB300,
    SUM(AB10_IND) AS AB10,
    SUM(AB20_IND) AS AB20,
    SUM(AB30_IND) AS AB30
FROM TABLE.VIEW.MAW
    UNPIVOT(PRODUCT_SELECTED FOR PRODUCT IN (AB4_IND, AB5_IND))
WHERE PRODUCT_SELECTED = 1
GROUP BY 1
;

【讨论】：

【解决方案3】：

union 通常比 OR 执行得更好，因为优化器通常难以获得正确的估计，但可以帮助您的一件事是通过使用子查询将操作数限制为一次：

SELECT
    Product,
    COUNT(*) AS CNT,
    SUM(CASE WHEN MULTI_UNIT = 1 THEN 1 ELSE 0 END) AS MULTI,
    SUM(AB300_IND) AS AB300,
    SUM(AB10_IND) AS AB10,
    SUM(AB20_IND) AS AB20,
    SUM(AB30_IND) AS AB30
from (  select 'AB4' AS Product,MULTI_UNIT,AB300_IND,AB10_IND,AB20_IND,AB30_IND
        FROM TABLE.VIEW.MAW
        WHERE AB4_IND = 1
        UNION ALL 
        select 'AB5',MULTI_UNIT,AB300_IND,AB10_IND,AB20_IND,AB30_IND
        FROM TABLE.VIEW.MAW
        WHERE AB5_IND = 1
) t group by product

【讨论】：