[转]Oracle存在则更新,不存在则插入

原文:http://hi.baidu.com/mawf2008/item/eec8c7ad1c5be5ae29ce9da6

merge into a 
using b
on (a.a=b.b)
when matched then update xxxxx
when not matched then insert (xxx) values(xxx);

oracle使用 merge 更新或插入数据(总结) 总结下。使用merge比传统的先判断再选择插入或更新快很多。
1)主要功能
提供有条件地更新和插入数据到数据库表中
如果该行存在,执行一个UPDATE操作,如果是一个新行,执行INSERT操作
— 避免了分开更新
— 提高性能并易于使用
— 在数据仓库应用中十分有用

2)MERGE语句的语法如下:

MERGE [hint] INTO [schema .] table [t_alias] USING [schema .]

{ table | view | subquery } [t_alias] ON ( condition )

WHEN MATCHED THEN merge_update_clause

WHEN NOT MATCHED THEN merge_insert_clause;


还是看例子就知道怎么回事:
MERGE INTO copy_emp c
USING employees e
ON (c.employee_id=e.employee_id)
WHEN MATCHED THEN
UPDATE SET
c.first_name=e.first_name,
c.last_name=e.last_name,
c.department_id=e.department_id
WHEN NOT MATCHED THEN
INSERT VALUES(e.employee_id,e.first_name,e.last_name,
e.email,e.phone_number,e.hire_date,e.job_id,
e.salary,e.commission_pct,e.manager_id,
e.departmetn_id);
MERGE INTO copy_emp c
USING employees e
ON (c.employee_id=e.employee_id)
WHEN MATCHED THEN
UPDATE SET
c.first_name=e.first_name,
c.last_name=e.last_name,
c.department_id=e.department_id
WHEN NOT MATCHED THEN
INSERT VALUES(e.employee_id,e.first_name,e.last_name,
e.email,e.phone_number,e.hire_date,e.job_id,
e.salary,e.commission_pct,e.manager_id,
e.departmetn_id);


3)使用merge的注意事项:
创建测试表:
CREATE TABLE MM (ID NUMBER, NAME VARCHAR2(20));
CREATE TABLE MN (ID NUMBER, NAME VARCHAR2(20));
插入数据
INSERT INTO MM VALUES (1, 'A');
INSERT INTO MN VALUES (1, 'B');
执行:
MERGE INTO MN A
USING MM B
ON(A.ID=B.ID)
WHEN MATCHED THEN
UPDATE SET A.ID = B.ID
WHEN NOT MATCHED THEN
INSERT VALUES(B.ID, B.NAME);
ON(A.ID=B.ID)
报错:无效的标识符,这个错误提示有些误导嫌疑,原因是on子句的使用的字段不能够用于update,即Oracle不允许更新用于连接 的列
修改:
MERGE INTO MN A
USING MM B
ON(A.ID=B.ID)
WHEN MATCHED THEN
UPDATE SET A.NAME = B.NAME
WHEN NOT MATCHED THEN
INSERT VALUES(B.ID, B.NAME);
ON(A.ID=B.ID)

再插入:INSERT INTO MM VALUES (1, 'C');
再执行:
MERGE INTO MN A
USING MM B
ON(A.ID=B.ID)
WHEN MATCHED THEN
UPDATE SET A.NAME = B.NAME
WHEN NOT MATCHED THEN
INSERT VALUES(B.ID, B.NAME);
ON(A.ID=B.ID)
报错,原因无法在源表中获得一组稳定的行

4)更新同一张表的数据。需要注意下细节,因为可能涉及到using的数据集为null,所以要使用count()函数。
MERGE INTO mn a
USING (select count(*) co from mn where mn.ID=4) b
ON (b.co<>0)--这里使用了count和<>,注意下,想下为什么!
WHEN MATCHED THEN
UPDATE
SET a.NAME = 'E'
where a.ID=4
WHEN NOT MATCHED THEN
INSERT
VALUES (4, 'E');

原文地址:https://www.cnblogs.com/dreamysmurf/p/3777904.html