如何改进包含存储过程使用的多个自联接的视图

Posted 2023-05-08

【中文标题】如何改进包含存储过程使用的多个自联接的视图

【英文标题】：How to improve a view containing multiple self joins used by a stored procedure

【发布时间】：2020-10-30 00:13:56

【问题描述】：

我有一个非常慢的存储过程（需要 5-6 分钟才能得到结果）包含一些表和一个视图。我相信是视图中的多个自连接部分使存储过程变慢。这里表 A 是一个 700,000 行的表，B 是一个 20 行的表。

表 A

id    | status_key | status_date | seq
10035       2         2020-10-01    1
10035       3         2020-10-03    2
10049       2         2020-06-10    1
10049       3         2020-06-13    2
10049       4         2020-06-17    3
10049       5         2020-07-03    4

表 B

status_key | status_name 
    2      | accepted
    3      | conditionally accepted
    4      | decided
    5      | declined

景色

SELECT a1.status_key as current_status_key,
       b1.status_name as current_status_name,
       a1.status_date as current_status_date,
       a2.status_key as previous_status_key,
       b2.status_name as precious_status_name,
       a2.status_date as previous_status_date,
       a3.status_key as next_status_key,
       b3.status_name as next_status_name,
       a3.status_date as next_status_date,
       a4.status_key as next_2_status_key,
       b4.status_name as next_2_status_name,
       a4.status_date as next_2_status_date,
FROM A a1
INNER JOIN B b1 ON a1.status_key = b1.status_key
LEFT JOIN A a2 ON a1.id = a2.id AND a1.seq = a2.seq + 1
LEFT JOIN B b2 ON a2.status_key = b2.status_key
LEFT JOIN A a3 ON a1.id = a3.id AND a1.seq = a3.seq - 1
LEFT JOIN B b3 ON a3.status_key = b3.status_key
LEFT JOIN A a4 ON a1.id = a4.id AND a1.seq = a4.seq - 2
LEFT JOIN B b4 ON a4.status_key = b4.status_key

想要的结果

 id  | current_status_key | current_status_name | current_status_date | previous_status_key | previous_status_name | previous_status_date | next_status_key | next_status_name | next_status_date | next_2_status_key | next_2_status_name | next_2_status_date 
10035 | 2 | accepted | 2020-10-01 | NULL | NULL | NULL | 3 | conditionally accepted | 2020-10-03 | NULL | NULL | NULL
10035 | 3 | conditionally accepted | 2020-10-03 | 2 | accepted | 2020-10-01 | | NULL | NULL | NULL | NULL | NULL | NULL

如何通过重写这部分来改善我的观点？我正在考虑使用 CTE 来分离上述部分。有什么想法吗？

【问题讨论】：

请提供样本数据、期望的结果以及您想要达到的目标的说明。

程序代码是高度特定于供应商的 - 所以请添加一个标签来指定您是否使用mysql、postgresql、sql-server、oracle或db2 - 或完全不同的东西。

【参考方案1】：

您可以改用 LAG 和 LEAD 来计算相关的状态键，而不是 A 上的自联接。这希望这意味着它只需要从 A 读取行一次 - 但需要根据您的数据库进行测试/等等

这是上面的 SQL 示例。注意 - SQL 已在问题更新提供数据后更新。

WITH a1 AS
    (SELECT A.ID,
       A.status_key AS a1_status_key,
       A.status_date AS a1_status_date,
       LAG(A.status_key, 1) OVER (PARTITION BY A.id ORDER BY A.seq) AS a2_status_key,
       LAG(A.status_date, 1) OVER (PARTITION BY A.id ORDER BY A.seq) AS a2_status_date,
       LEAD(A.status_key, 1) OVER (PARTITION BY A.id ORDER BY A.seq) AS a3_status_key,
       LEAD(A.status_date, 1) OVER (PARTITION BY A.id ORDER BY A.seq) AS a3_status_date,
       LEAD(A.status_key, 2) OVER (PARTITION BY A.id ORDER BY A.seq) AS a4_status_key,
       LEAD(A.status_date, 2) OVER (PARTITION BY A.id ORDER BY A.seq) AS a4_status_date
     FROM A
    ) 
SELECT a1.id, 
       a1.a1_status_key as current_status_key,
       b1.status_name as current_status_name,
       a1.a1_status_date as current_status_date,
       a1.a2_status_key as previous_status_key,
       b2.status_name as previous_status_name,
       a1.a2_status_date as previous_status_date,
       a1.a3_status_key as next_status_key,
       b3.status_name as next_status_name,
       a1.a3_status_date as next_status_date,
       a1.a4_status_key as next_2_status_key,
       b4.status_name as next_2_status_name,
       a1.a4_status_date as next_2_status_date
FROM a1
    LEFT JOIN B b1 ON a1.a1_status_key = b1.status_key
    LEFT JOIN B b2 ON a1.a2_status_key = b2.status_key
    LEFT JOIN B b3 ON a1.a3_status_key = b3.status_key
    LEFT JOIN B b4 ON a1.a4_status_key = b4.status_key;

这是一个使用临时表的db<>fiddle。

如果聚集索引位于id, seq 上，我认为它也会有很大帮助。如果表 A 实际上更大并且具有其他值，则在这两列上使用非聚集索引，然后包含其他相关列可能会更好，例如 id, seq, status_date, status_key。

以前的版本

WITH a1 AS
    (SELECT ...,
       LEAD(A.status_key, 1) OVER (PARTITION BY A.id ORDER BY A.seq) AS a2_status_key,
       LAG(A.status_key, 1) OVER (PARTITION BY A.id ORDER BY A.seq) AS a3_status_key,
       LAG(A.status_key, 2) OVER (PARTITION BY A.id ORDER BY A.seq) AS a4_status_key
     FROM A
    ) 
SELECT a1.*, ...
FROM a1
    LEFT JOIN B b1 ON a1.status_key = b1.status_key
    LEFT JOIN B b2 ON a1.a2_status_key = b2.status_key
    LEFT JOIN B b3 ON a1.a3_status_key = b3.status_key
    LEFT JOIN B b4 ON a1.a4_status_key = b4.status_key;

【讨论】：

如果这些点使用了来自另一个 seqnr 的大量列，那么将会有很多 LAGs，我希望 DBMS 对每个偏移值只执行一次偏移。另一种选择是物化视图而不是普通视图，如果多次访问此视图肯定会提高性能，但会占用额外的空间。无论如何，对于这种顺序项目的情况，您的解决方案非常好。

你是对的。我使用的是基于他们使用SELECT ... 的 OP（预编辑）问题的点 - 所以我只是使用相同的，但如果他们从 a2、a3 和 a4 中选择很多列，那么是的，这将是很多领先/滞后。从他们的更新来看，他们似乎每个（日期和状态键）都使用了 2 个。我认为值得一试性能，尽管索引也会有所帮助。

谢谢肖恩。我使用了您的方法并用数据对其进行了测试。它真的奏效了！现在只需 20 秒运行存储过程即可返回前 50 行！