merge应用一例

如何把id 为1和2的两列的值互换？

scott@LS1>create table test(id number,name varchar2(20));

Table created.

scott@LS1>insert into test values(1,'a');

1 row created.

scott@LS1>insert into test values(2,'b');

1 row created.

scott@LS1>commit;

Commit complete.

scott@LS1>select * from test;

ID NAME

---------- --------------------

1 a

2 b

当你执行update test set name =（select name from test where id =2 ) where id =1;

这个时候id=1 的那么的值已经改变了，不能实现目标

思路一：

所以我们需要构建一个虚表，把原来的值存储起来，然后和原表merge

scott@LS1>select 1 id,(select name from test where id=2) name from dual

2 union all

3 select 2,(select name from test where id=1) from dual;

ID NAME

---------- --------------------

1 b

2 a

scott@LS1>merge into test

2 using (select 1 id ,(select name from test where id=2) name from dual

3 union all

4 select 2,(select name from test where id=1) from dual) t

5 on (test.id = t.id)

6 when matched then update set test.name = t.name;

2 rows merged.

scott@LS1>select * from test;

ID NAME

---------- --------------------

1 b

2 a

思路二：

直接把真实结果用select 的方式取出，然后利用这个结果集更新原表记录。

scott@LS1>l

1 merge into test using

2 (

3 with t as

4 (select 1 id,(select name from test where id=2 ) name from dual

5 union all

6 select 2,(select name from test where id=1 ) from dual)

7 select test.id,test.rowid as rn,t.name from test,t

8 where test.id = t.id

9 )n

10 on(test.rowid = n.rn)

11 when matched then update

12* set test.name = n.name

scott@LS1>/

2 rows merged.

scott@LS1>select * from test;

ID NAME

---------- --------------------

1 b

2 a

仅仅是查询的话，可以使用：

scott@LS1>with t as

2 (select 1 id,(select name from test where id =2) name from dual

3 union all

4 select 2, (select name from test where id =1) from dual)

5 select test.id, t.name from test,t

6 where test.id = t.id;

ID NAME

---------- --------------------

1 b

2 a