一、业务场景：
（1）主从两个表，主表Student，有字段id、name、sex，从表Boy，有字段id、name，主从表同一对象id相同
（2）从表Boy的name属性被业务修改，定时批量处理主表，以维持主表name属性与从表一致
二、表结构
1、主表 Student
2、从表 Boy
三、建表SQL（DDL）
1、主表 Student
1 -- DDL 2 CREATE TABLE student ( 3 id NUMBER NOT NULL , 4 name VARCHAR2(255 BYTE) NULL , 5 sex VARCHAR2(255 BYTE) NULL 6 ) 7 8 ALTER TABLE student ADD CHECK (id IS NOT NULL); 9 10 -- DML 11 INSERT INTO student VALUES ('1', 'zhangsan', 'boy'); 12 INSERT INTO student VALUES ('2', 'lisi', 'girl'); 13 INSERT INTO student VALUES ('3', 'wangwu', 'boy');
2、从表 Boy
1 -- DDL 2 CREATE TABLE boy ( 3 id NUMBER NOT NULL , 4 name VARCHAR2(255 BYTE) NULL 5 ) 6 7 -- DML 8 INSERT INTO boy VALUES ('1', '张三'); 9 INSERT INTO boy VALUES ('3', '王五');
四、DML
1、基本语法
1 -- DML 2 UPDATE student s SET s.name = '张三' WHERE id = 1;
2、变相
1 -- DML，0.015s 2 UPDATE student s 3 SET s.name = ( 4 SELECT b.name FROM boy b WHERE s.id = b.id AND s.name != b.name 5 ) 6 WHERE EXISTS ( 7 SELECT 1 FROM boy b WHERE s.id = b.id AND s.name != b.name 8 ); 9
3、快速游标法
1 -- DML，0.014s 2 BEGIN 3 FOR cur IN ( 4 SELECT s.id sid, b.name bname 5 FROM student s, boy b 6 WHERE s.id = b.id AND s.name != b.name AND s.sex = 'boy' 7 ) loop 8 9 UPDATE student s SET s.name = cur.bname WHERE s.id = cur.sid; 10 11 END loop ; 12 END ; 13
4、内联视图法（inline View）
1 -- DML，0.019s 2 UPDATE ( 3 SELECT 4 s.name sname, b.name bname 5 FROM 6 student s, boy b 7 WHERE 8 s.id = b.id AND s.name != b.name 9 ) 10 SET sname = bname; 11
报错提示：ORA-01779: 无法修改与非键值保存表对应的列
参考资料，从表id必须增加主键约束，且为视图内的where条件：
1 -- DDL 2 ALTER TABLE boy ADD CONSTRAINT pk_id PRIMARY KEY (id);
5、合并法（Merge）
1 MERGE INTO student s USING boy b ON ( 2 s.id = b.id AND s.sex = 'boy' AND s.name != b.name 3 ) 4 WHEN MATCHED THEN 5 UPDATE SET s.name = b.name;
报错提示：ORA-38104: 无法更新 ON 子句中引用的列
参考资料，错误原因是条件重复，正确写法：
1 -- DML，0.016s 2 MERGE INTO student s USING boy b ON ( 3 s.id = b.id AND s.sex = 'boy' 4 -- AND s.name != b.name 5 ) 6 WHEN MATCHED THEN 7 UPDATE SET s.name = b.name;
6、子查询关联
1 update service cs set (cs.customer_id,cs.customer_code,cs.customer_name)= 2 ( 3 select ccb.customer_id,ccb.customer_code,ccb.customer_name from customer_base ccb 4 inner join customer_contact ccc 5 on ccb.customer_id=ccc.customer_id 6 where ccc.key_='1' and ccc.value_=cs.value_ and rownum<2 7 ) 8 where customer_id='0' and start_time>sysdate-30