Question

在以下原始查询中：

SELECT COMPANYNAME,
    (
        SELECT SUM(RRP) * 0.1
        FROM CRM_RESALE_ITEM_VIEW
        INNER JOIN CRM_RESALE using (RESALE_ID)
        WHERE CRM_RESALE.CUSTOMER_ID = CRM_CUSTOMER_VIEW.CUSTOMER_ID
        ) AS DERRIVED_MAINTENANCE
FROM CRM_CUSTOMER_VIEW

我已将DERRIVED_MAINTENANCE子选项替换为如下：

SELECT COMPANYNAME,
    F_MAINTENANCE(CRM_CUSTOMER_VIEW.CUSTOMER_ID) AS DERRIVED_MAINTENANCE
FROM CRM_CUSTOMER_VIEW

有一个功能：

BEGIN
    DECLARE DERRIVED_MAINTENANCE DECIMAL DEFAULT 0;

    SELECT SUM(RRP) * 0.1
    INTO DERRIVED_MAINTENANCE
    FROM CRM_RESALE_ITEM_VIEW
    INNER JOIN CRM_RESALE using (RESALE_ID)
    WHERE CRM_RESALE.CUSTOMER_ID = CUST_ID;

    RETURN DERRIVED_MAINTENANCE;
END

现在，而不是花费60秒，查询永远不会返回。任何人都可以看到这个原因吗？

CRM_CUSTOMER (CUSTOMER_ID) one-to-many with 
CRM_RESALE (RESALE_ID, CUSTOMER_ID) one-to-many with 
CRM_RESALE_ITEM_VIEW (RESALE_ID, ITEM_ID, RRP)

Answer 1

TL; DR ：通过正确的索引可以为您提供更好的服务。将JOIN卸载到函数实际上可能会使事情变得更糟，更糟糕。但是正确索引VIEW并不简单，并且您没有提供足够的信息来提供有保证的解决方案。您可以在下面找到提案和评估的测试。

可能是函数返回，但它需要很长时间。我在与原始查询大致相同的时间内返回（我的定义与您的定义不同，仔细检查我的代码，我可能会误解）：

示例数据

CREATE TABLE CRM_CUSTOMER_VIEW
    ( CUSTOMER_ID INTEGER, COMPANYNAME VARCHAR(50) );

INSERT INTO CRM_CUSTOMER_VIEW VALUES ( 1, 'ACME' ), ( 2, 'NASA' );

SELECT @N:=COUNT(*) FROM CRM_CUSTOMER_VIEW;
INSERT INTO CRM_CUSTOMER_VIEW SELECT CUSTOMER_ID + @N, CONCAT(SUBSTRING(COMPANYNAME, 1, 4), ' ', @N, '.', CUSTOMER_ID)
    FROM CRM_CUSTOMER_VIEW;
-- Repeat the two rows above to fill the table with, say, half a million records.

CREATE TABLE CRM_RESALE ( CUSTOMER_ID INTEGER, RESALE_ID INTEGER );    
SELECT @N:=1;

INSERT INTO CRM_RESALE SELECT CUSTOMER_ID, 5*CUSTOMER_ID+@N FROM CRM_CUSTOMER_VIEW;
SELECT @N:=@N+1;
-- Repeat five times the two rows above to get five items per customer

CREATE TABLE CRM_RESALE_ITEM_VIEW ( RESALE_ID INTEGER, RRP NUMERIC(7,3));
INSERT INTO CRM_RESALE_ITEM_VIEW SELECT RESALE_ID, 3.14159 FROM CRM_RESALE;

现在我们运行查询以获得基线 - 没有索引，因此即使在快速机器上也非常昂贵

SELECT COMPANYNAME,
    (
        SELECT SUM(RRP) * 0.1
        FROM CRM_RESALE_ITEM_VIEW
        INNER JOIN CRM_RESALE using (RESALE_ID)
        WHERE CRM_RESALE.CUSTOMER_ID = CRM_CUSTOMER_VIEW.CUSTOMER_ID
        ) AS DERRIVED_MAINTENANCE
FROM CRM_CUSTOMER_VIEW WHERE COMPANYNAME = 'ACME';

+-------------+----------------------+
| COMPANYNAME | DERRIVED_MAINTENANCE |
+-------------+----------------------+
| ACME        |               1.5710 |
+-------------+----------------------+
1 row in set (3.18 sec)

现在我们将内部查询移动到它自己的函数中。

DELIMITER //
CREATE FUNCTION DERRIVED_MAINTENANCE ( CUSTID INTEGER )
RETURNS NUMERIC(7,3) DETERMINISTIC
BEGIN
SELECT SUM(RRP) * 0.1 INTO @SRRP
        FROM CRM_RESALE_ITEM_VIEW
        INNER JOIN CRM_RESALE using (RESALE_ID)
        WHERE CRM_RESALE.CUSTOMER_ID = CUSTID;

RETURN @SRRP;
END//
DELIMITER ;

mysql> SELECT DERRIVED_MAINTENANCE(1);
+-------------------------+
| DERRIVED_MAINTENANCE(1) |
+-------------------------+
|                   1.571 |
+-------------------------+
1 row in set (3.60 sec)

如果我针对FIVE行运行查询，我会五次，因为该函数被调用了五次。

SELECT CUSTOMER_ID, DERRIVED_MAINTENANCE(CUSTOMER_ID) FROM CRM_CUSTOMER_VIEW WHERE CUSTOMER_ID < 5;
+-------------+-----------------------------------+
| CUSTOMER_ID | DERRIVED_MAINTENANCE(CUSTOMER_ID) |
+-------------+-----------------------------------+
|           1 |                             1.571 |
|           2 |                             1.571 |
|           3 |                             1.571 |
|           4 |                             1.571 |
+-------------+-----------------------------------+
4 rows in set (14.45 sec)

如果我索引这些表，但使用覆盖索引 - 我可以这样做，因为它们是表格;您也可以为视图执行此操作，但您需要以不同的方式编制索引，并且可能您可以从不同的聚合视图中受益。如果不了解更多，就无法真正建议

CREATE INDEX CRM_RESALE_ITEM_VIEW_NDX ON CRM_RESALE_ITEM_VIEW(RESALE_ID, RRP);
CREATE INDEX CRM_RESALE_NDX ON CRM_RESALE (CUSTOMER_ID, RESALE_ID);

现在我可以使用VIEW代替函数，或者调用函数：

CREATE VIEW CRM_RESALE_FULL_VIEW AS
        SELECT CUSTOMER_ID, SUM(RRP) * 0.1 AS DERRIVED_MAINTENANCE
        FROM CRM_RESALE_ITEM_VIEW
        INNER JOIN CRM_RESALE using (RESALE_ID)
    GROUP BY CUSTOMER_ID;

SELECT COMPANYNAME, DERRIVED_MAINTENANCE FROM     CRM_CUSTOMER_VIEW JOIN CRM_RESALE_FULL_VIEW USING (CUSTOMER_ID)     WHERE COMPANYNAME LIKE 'ACME 1024.20%';
+---------------+----------------------+
| COMPANYNAME   | DERRIVED_MAINTENANCE |
+---------------+----------------------+
| ACME 1024.201 |               1.5710 |
| ACME 1024.203 |               1.5710 |
| ACME 1024.205 |               1.5710 |
| ACME 1024.207 |               1.5710 |
| ACME 1024.209 |               1.5710 |
+---------------+----------------------+
5 rows in set (1.11 sec)

SELECT COMPANYNAME, DERRIVED_MAINTENANCE FROM     CRM_CUSTOMER_VIEW JOIN CRM_RESALE_FULL_VIEW USING (CUSTOMER_ID)     WHERE COMPANYNAME LIKE 'ACME 1024.2%';
+---------------+----------------------+
| COMPANYNAME   | DERRIVED_MAINTENANCE |
+---------------+----------------------+
| ACME 1024.21  |               1.5710 |
...
| ACME 1024.299 |               1.5710 |
+---------------+----------------------+
55 rows in set (1.31 sec)

或调用函数

SELECT CUSTOMER_ID, DERRIVED_MAINTENANCE(CUSTOMER_ID) FROM CRM_CUSTOMER_VIEW WHERE CUSTOMER_ID > 42 AND CUSTOMER_ID < 47;

SELECT CUSTOMER_ID, DERRIVED_MAINTENANCE(CUSTOMER_ID) FROM CRM_CUSTOMER_VIEW WHERE CUSTOMER_ID > 42 AND CUSTOMER_ID < 47;
+-------------+-----------------------------------+
| CUSTOMER_ID | DERRIVED_MAINTENANCE(CUSTOMER_ID) |
+-------------+-----------------------------------+
|          43 |                             1.571 |
|          44 |                             1.571 |
|          45 |                             1.571 |
|          46 |                             1.571 |
+-------------+-----------------------------------+
4 rows in set (0.03 sec)

单独建立索引可以使性能提升两个数量级。

mysql用函数替换子选择 - 查询永远不会返回

1 个答案: