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Question

我创建了这个视图，它的查询速度非常慢，并且会产生性能问题。我认为我需要使用 PIVOT，但我无法真正理解它......

这是我目前所拥有的。

select
    p.instrumentsettingsid,
    sum(case when p1.id is null then 0 else 1 end) as 'Cash',
    sum(case when p2.id is null then 0 else 1 end) as HighInterest,
    sum(case when p3.id is null then 0 else 1 end) as 'ML Loan',
    sum(case when p4.id is null then 0 else 1 end) as 'Investment Note',
    sum(case when p5.id is null then 0 else 1 end) as 'Trading',
    sum(case when p6.id is null then 0 else 1 end) as 'DD Trading',
    sum(case when p7.id is null then 0 else 1 end) as 'ML Trading',
    sum(case when p8.id is null then 0 else 1 end) as 'NMLTrading',
    sum(case when p9.id is null then 0 else 1 end) as 'Individual Custody',
    sum(case when p10.id is null then 0 else 1 end) as 'ML Individual Custody',
    sum(case when p11.id is null then 0 else 1 end) as 'Custody',
    sum(case when p12.id is null then 0 else 1 end) as 'ML Custody Trading',
    sum(case when p13.id is null then 0 else 1 end) as 'Portfolio',
    sum(case when p14.id is null then 0 else 1 end) as 'ML Personal Portfolio',
    sum(case when p15.id is null then 0 else 1 end) as Other,
    i.status
from
    (select distinct instrumentsettingsid from res_db..instrumentproducttypepermissions) p
    inner join res_db..instrumentsettings i on p.instrumentsettingsid = i.id
    left join res_db..instrumentproducttypepermissions p1 on p.instrumentsettingsid = p1.instrumentsettingsid and p1.enabled = 1 and p1.producttypeid = 1
    left join res_db..instrumentproducttypepermissions p2 on p.instrumentsettingsid = p2.instrumentsettingsid and p2.enabled = 1 and p2.producttypeid = 2
    left join res_db..instrumentproducttypepermissions p3 on p.instrumentsettingsid = p3.instrumentsettingsid and p3.enabled = 1 and p3.producttypeid = 3
    left join res_db..instrumentproducttypepermissions p4 on p.instrumentsettingsid = p4.instrumentsettingsid and p4.enabled = 1 and p4.producttypeid = 4
    left join res_db..instrumentproducttypepermissions p5 on p.instrumentsettingsid = p5.instrumentsettingsid and p5.enabled = 1 and p5.producttypeid = 5
    left join res_db..instrumentproducttypepermissions p6 on p.instrumentsettingsid = p6.instrumentsettingsid and p6.enabled = 1 and p6.producttypeid = 6
    left join res_db..instrumentproducttypepermissions p7 on p.instrumentsettingsid = p7.instrumentsettingsid and p7.enabled = 1 and p7.producttypeid = 7
    left join res_db..instrumentproducttypepermissions p8 on p.instrumentsettingsid = p8.instrumentsettingsid and p8.enabled = 1 and p8.producttypeid = 8
    left join res_db..instrumentproducttypepermissions p9 on p.instrumentsettingsid = p9.instrumentsettingsid and p9.enabled = 1 and p9.producttypeid = 9
    left join res_db..instrumentproducttypepermissions p10 on p.instrumentsettingsid = p10.instrumentsettingsid and p10.enabled = 1 and p10.producttypeid = 10
    left join res_db..instrumentproducttypepermissions p11 on p.instrumentsettingsid = p11.instrumentsettingsid and p11.enabled = 1 and p11.producttypeid = 11
    left join res_db..instrumentproducttypepermissions p12 on p.instrumentsettingsid = p12.instrumentsettingsid and p12.enabled = 1 and p12.producttypeid = 12
    left join res_db..instrumentproducttypepermissions p13 on p.instrumentsettingsid = p13.instrumentsettingsid and p13.enabled = 1 and p13.producttypeid = 13
    left join res_db..instrumentproducttypepermissions p14 on p.instrumentsettingsid = p14.instrumentsettingsid and p14.enabled = 1 and p14.producttypeid = 14
    left join res_db..instrumentproducttypepermissions p15 on p.instrumentsettingsid = p15.instrumentsettingsid and p15.enabled = 1 and p15.producttypeid = 15
    group by p.instrumentsettingsid, i.status

知道 SQL 的人可以告诉我（告诉我吗？）我怎样才能使它更快/不那么可怕。

此外，这 15 个帐户名称是硬编码的，在理想情况下，它会从表 producttypes 中获取产品名称

很抱歉问了这么烦人的问题。

谢谢！

Answer 1

用途：

  SELECT p.instrumentsettingsid,
         i.status,
         SUM(CASE WHEN p.producttypeid = 1 THEN 1 ELSE 0 END) as 'Cash',
         SUM(CASE WHEN p.producttypeid = 2 THEN 1 ELSE 0 END) as HighInterest,
         SUM(CASE WHEN p.producttypeid = 3 THEN 1 ELSE 0 END) as 'ML Loan',
         SUM(CASE WHEN p.producttypeid = 4 THEN 1 ELSE 0 END) as 'Investment Note',
         SUM(CASE WHEN p.producttypeid = 5 THEN 1 ELSE 0 END) as 'Trading',
         SUM(CASE WHEN p.producttypeid = 6 THEN 1 ELSE 0 END) as 'DD Trading',
         SUM(CASE WHEN p.producttypeid = 7 THEN 1 ELSE 0 END) as 'ML Trading',
         SUM(CASE WHEN p.producttypeid = 8 THEN 1 ELSE 0 END) as 'NMLTrading',
         SUM(CASE WHEN p.producttypeid = 9 THEN 1 ELSE 0 END) as 'Individual Custody',
         SUM(CASE WHEN p.producttypeid = 10 THEN 1 ELSE 0 END) as 'ML Individual Custody',
         SUM(CASE WHEN p.producttypeid = 11 THEN 1 ELSE 0 END) as 'Custody',
         SUM(CASE WHEN p.producttypeid = 12 THEN 1 ELSE 0 END) as 'ML Custody Trading',
         SUM(CASE WHEN p.producttypeid = 13 THEN 1 ELSE 0 END) as 'Portfolio',
         SUM(CASE WHEN p.producttypeid = 14 THEN 1 ELSE 0 END) as 'ML Personal Portfolio',
         SUM(CASE WHEN p.producttypeid = 15 THEN 1 ELSE 0 END) as Other,
    FROM res_db..instrumentproducttypepermissions p
    JOIN res_db..instrumentsettings i ON i.id = p.instrumentsettingsid
   WHERE p.enabled = 1
     AND p.producttypeid BETWEEN 1 AND 15
GROUP BY p.instrumentsettingsid, i.status

不需要左连接...

另外，这 15 个帐户名称是硬编码的，在理想情况下，它会从表 producttypes 中获取产品名称

动态列需要dynamic SQL。

Answer 2

即使使用 Sql Server PIVOT，您也必须事先知道列名，所以没有运气，除非您使用 Pivots with Dynamic Columns 或 Dynamic PIVOT