在 Oracle SQL 查询中避免不必要的连接

Question

假设你有一个如下表：

表名：客户 主键：CUSTOMER_ID +-------------+--------------+ |客户 ID |客户名称 | +-------------+--------------+ | 1 |比尔 | | 2 |汤姆 | +-------------+--------------+

现在，假设您有一个 CUSTOMER_ATTRIBUTE 表，可让您将键/值对绑定到特定的 CUSTOMER：

表名：CUSTOMER_ATTRIBUTE 主键：（CUSTOMER_ID，ATTRIBUTE_TYPE_ID） +-------------+-------+-------------- --+ |客户 ID | ATTRIBUTE_TYPE_ID | ATTRIBUTE_VALUE | +-------------+-------+-------------- --+ | 1 | FAVORITE_FOOD |披萨 | | 1 | FAVORITE_COLOR |蓝色 | | 2 | FAVORITE_FOOD |炸玉米饼 | | 2 | NAME_OF_PET |菲多 | +-------------+-------+-------------- --+

现在，假设您创建了一个视图来表示具有某些可能属性的客户：

CREATE VIEW CUSTOMER_VIEW AS
SELECT
    CUSTOMER.CUSTOMER_ID,
    CUSTOMER.CUSTOMER_NAME,
    FAVORITE_FOOD_ATTRIBUTE.ATTRIBUTE_VALUE AS FAVORITE_FOOD,
    FAVORITE_COLOR_ATTRIBUTE.ATTRIBUTE_VALUE AS FAVORITE_COLOR
FROM
    CUSTOMER

    LEFT OUTER JOIN CUSTOMER_ATTRIBUTE favorite_food_attribute
        ON customer.customer_id = favorite_food_attribute.customer_id
           AND favorite_food_attribute.attribute_type_id = FAVORITE_FOOD

    LEFT OUTER JOIN CUSTOMER_ATTRIBUTE favorite_color_attribute
        ON customer.customer_id = favorite_color_attribute.customer_id
           AND favorite_color_attribute.attribute_type_id = FAVORITE_COLOR

现在，假设您查询这个视图：

SELECT
    CUSTOMER_ID,
    CUSTOMER_NAME,
    FAVORITE_COLOR
    -- Notice: I did not ask for the FAVORITE_FOOD column
FROM
    CUSTOMER_VIEW

根据解释计划，Oracle 仍在加入favorite_food_attribute，尽管它的值不需要并且不影响查询的基数（因为它是LEFT OUTER JOIN 对表的主键）。

有没有办法强制 Oracle 避免这些不必要的连接？

更新：DDL 示例

这里是一些用于创建示例架构的 DDL：

CREATE TABLE CUSTOMER
(
    CUSTOMER_ID   NUMBER NOT NULL,
    CUSTOMER_NAME VARCHAR2(100)
);

CREATE UNIQUE INDEX CUSTOMER_PK_INDEX
    ON CUSTOMER(CUSTOMER_ID);

ALTER TABLE CUSTOMER
    ADD CONSTRAINT CUSTOMER_PK
    PRIMARY KEY (CUSTOMER_ID)
    USING INDEX CUSTOMER_PK_INDEX;

CREATE TABLE CUSTOMER_ATTRIBUTE
(
    CUSTOMER_ID       NUMBER NOT NULL,
    ATTRIBUTE_TYPE_ID NUMBER NOT NULL,
    ATTRIBUTE_VALUE   VARCHAR2(1000)
);

CREATE UNIQUE INDEX CUSTOMER_ATTRIBUTE_PK_INDEX
    ON CUSTOMER_ATTRIBUTE(CUSTOMER_ID, ATTRIBUTE_TYPE_ID);

ALTER TABLE CUSTOMER_ATTRIBUTE
    ADD CONSTRAINT CUSTOMER_ATTRIBUTE_PK
    PRIMARY KEY (CUSTOMER_ID, ATTRIBUTE_TYPE_ID)
    USING INDEX CUSTOMER_ATTRIBUTE_PK_INDEX;

CREATE OR REPLACE VIEW CUSTOMER_VIEW AS
SELECT
    CUSTOMER.CUSTOMER_ID,
    CUSTOMER.CUSTOMER_NAME,
    favorite_food_attribute.attribute_value AS favorite_food,
    favorite_color_attribute.attribute_value AS favorite_color
FROM
    CUSTOMER

    LEFT OUTER JOIN CUSTOMER_ATTRIBUTE favorite_food_attribute
        ON customer.customer_id = favorite_food_attribute.customer_id
           AND favorite_food_attribute.attribute_type_id = 5

    LEFT OUTER JOIN CUSTOMER_ATTRIBUTE favorite_color_attribute
        ON customer.customer_id = favorite_color_attribute.customer_id
           AND favorite_color_attribute.attribute_type_id = 6;

现在，我对这个查询运行解释计划：

SELECT CUSTOMER_ID FROM HFSMMM.CUSTOMER_VIEW

计划是：

SELECT STATEMENT, GOAL = ALL_ROWS 成本=1 基数=1 字节=65 嵌套循环外部成本=1 基数=1 字节=65 嵌套循环外部成本=1 基数=1 字节=39 INDEX FULL SCAN 对象所有者=HFSMMM 对象名称=CUSTOMER_PK_INDEX 成本=1 基数=1 字节=13 INDEX UNIQUE SCAN 对象所有者=HFSMMM 对象名称=CUSTOMER_ATTRIBUTE_PK_INDEX 成本=0 基数=1 字节=26 INDEX UNIQUE SCAN 对象所有者=HFSMMM 对象名称=CUSTOMER_ATTRIBUTE_PK_INDEX 成本=0 基数=1 字节=26

Answer 1

虽然这种方法只是移动处理而不是消除它，但它使 SQL 更清晰。
创建用户函数

GET_TYPE(customer_id_in NUMBER, attribute_type_id IN NUMBER) RETURN VARCHAR 2
IS
/*  TO DO:  Assertions, error handling */
attribute_name VARCHAR2(300);
BEGIN
SELECT attribute_value
INTO attribute_name
FROM CUSTOMER_ATTRIBUTE
WHERE customer_id = customer_id_in
and attribute_type_id - attribute_type_in;


RETURN attribute_name;

END GET_TYPE;

然后你的观点是

CREATE VIEW CUSTOMER_VIEW as
SELECT  
    CUSTOMER.CUSTOMER_ID,  
    CUSTOMER.CUSTOMER_NAME,  
    GET_TYPE(1, CUSTOMER.CUSTOMER_ID) AS FOOD,
    GET_TYPE(2, CUSTOMER.CUSTOMER_ID) AS COLOR
FROM  
    CUSTOMER;

Adam 指出切换上下文存在开销是正确的 我每天都用它来查看。我宁愿让数据库提前完成工作以准备视图和查询，而不是让应用程序发送必须构建和缓存的多连接查询。

Answer 2

如果您确定每个客户 ID 和属性类型只会有一个条目，则可以执行标量子查询：

SELECT
    CUSTOMER.CUSTOMER_ID,
    CUSTOMER.CUSTOMER_NAME,
    (select ATTRIBUTE_VALUE from CUSTOMER_ATTRIBUTE where customer_id = CUSTOMER.CUSTOMER_ID
        and ATTRIBUTE_TYPE_ID='F') AS FAVORITE_FOOD
FROM
    CUSTOMER

Answer 3

不要使用外连接，而是对您希望在视图中看到的每个属性值使用子查询。这是假设您的数据是结构化的，因此没有一个子查询可以返回多行。

CREATE VIEW CUSTOMER_VIEW AS
SELECT CUSTOMER_ID,
       CUSTOMER_NAME,
       (SELECT ATTRIBUTE_VALUE FROM CUSTOMER_ATTRIBUTE ca1
          WHERE ca1.CUSTOMER_ID = c.CUSTOMER_ID
          AND ATTRIBUTE_TYPE_ID = 'FAVFOOD')  FAVORITE_FOOD,
       (SELECT ATTRIBUTE_VALUE FROM CUSTOMER_ATTRIBUTE ca2
          WHERE ca2.CUSTOMER_ID = c.CUSTOMER_ID
          AND ATTRIBUTE_TYPE_ID = 'PETNAME')  PET_NAME,
       (SELECT ATTRIBUTE_VALUE FROM CUSTOMER_ATTRIBUTE ca3
          WHERE ca3.CUSTOMER_ID = c.CUSTOMER_ID
          AND ATTRIBUTE_TYPE_ID = 'FAVCOLOR') FAVORITE_COLOR
       FROM CUSTOMER c