Oracle批量、大量Update方法总结

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

-- DDL
CREATE TABLE student (
id NUMBER NOT NULL ,
name VARCHAR2(255 BYTE) NULL ,
sex VARCHAR2(255 BYTE) NULL
)
 
ALTER TABLE student ADD CHECK (id IS NOT NULL);
 
-- DML
INSERT INTO student VALUES ('1', 'zhangsan', 'boy');
INSERT INTO student VALUES ('2', 'lisi', 'girl');
INSERT INTO student VALUES ('3', 'wangwu', 'boy');

 2、从表 Boy

-- DDL
CREATE TABLE boy (
id NUMBER NOT NULL ,
name VARCHAR2(255 BYTE) NULL
)
 
-- DML
INSERT INTO boy VALUES ('1', '张三');
INSERT INTO boy VALUES ('3', '王五');

四、DML

  1、基本语法

-- DML
UPDATE student s SET s.name = '张三' WHERE   id = 1; 

  2、变相

 -- DML，0.015s
UPDATE student s
SET s.name = (
  SELECT b.name FROM boy b WHERE s.id = b.id AND s.name != b.name
)
WHERE EXISTS (
  SELECT 1 FROM boy b WHERE s.id = b.id AND s.name != b.name
);
 

3、快速游标法

 -- DML，0.014s
BEGIN
  FOR cur IN (
    SELECT s.id sid, b.name bname
  FROM student s, boy b
  WHERE s.id = b.id AND s.name != b.name AND s.sex = 'boy'
  ) loop 
 
  UPDATE student s SET s.name = cur.bname WHERE s.id = cur.sid;
  
  END loop ;
END ;
 

4、内联视图法（inline View）

 -- DML，0.019s
UPDATE (
  SELECT
    s.name sname, b.name bname
 FROM
    student s, boy b
 WHERE
    s.id = b.id AND s.name != b.name
)
SET sname = bname;
 

  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：

 -- DDL
ALTER TABLE boy ADD CONSTRAINT pk_id PRIMARY KEY (id);  

  5、合并法（Merge）

MERGE INTO student s USING boy b ON (
 s.id = b.id AND s.sex = 'boy' AND s.name != b.name
)
WHEN MATCHED THEN
 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：

 -- DML，0.016s
MERGE INTO student s USING boy b ON (
 s.id = b.id AND s.sex = 'boy'
--   AND s.name != b.name
)
WHEN MATCHED THEN
 UPDATE SET s.name = b.name;  

6、子查询关联

update service cs set (cs.customer_id,cs.customer_code,cs.customer_name)=  
 (
     select ccb.customer_id,ccb.customer_code,ccb.customer_name from customer_base ccb 
     inner join customer_contact ccc
     on ccb.customer_id=ccc.customer_id 
     where ccc.key_='1' and ccc.value_=cs.value_ and rownum<2
 )
 where customer_id='0' and start_time>sysdate-30