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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子选择替换如下：

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

现在查询不再需要 60 秒，而是永远不会返回。任何人都可以看到这样做的原因吗？

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)
4

1 回答 1

0

TL;DR：正确的索引可能会更好地为您服务。卸载JOIN到一个函数实际上可能会使事情变得更糟。但是正确索引 aVIEW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。下面，您将找到一个提案和一个要评估的测试。

可能是函数正在返回，但它需要很长时间。我让它与原始查询大约在同一时间返回（我的定义与您的不同，请仔细检查我的代码，我可能误解了）：

样本数据

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取基线 - 没有索引，因此非常昂贵，即使在快速机器上也是如此

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)

如果我对五行运行查询，我得到的时间是函数被调用五次后的五倍。

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)

但是，如果我使用覆盖索引对表进行索引——我可以这样做，因为它们是表；您也可以对视图执行此操作，但是您需要以不同的方式索引，也许您可​​以从不同的聚合视图中受益。在不了解更多信息的情况下无法真正提供建议

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 aVIEW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)

单独的索引可以产生两个数量级的性能提升。

于 2013-10-29T13:04:03.307 回答