merge

从ORACLE 9i开始，ORACLE提供了可以同时执行 insert ,update的命令merge 。

创建测试表 info

SQL> create table info(name varchar2(10),age number, address varchar2(10));

Table created.
插入3条测试数据

insert into info values('robinson',100,'chongqing');
insert into info values('luobingsen',100,'chongqing');
insert into info values('luoluo',100,'chongqing');

创建测试表 information

SQL>
create table information(name varchar2(10),id number,age number,address varchar2(10),school varchar2(10));
Table created.
插入4条测试数据

insert into information values('robinson',1,21,'chongqing','xinandaxue');
insert into information values('luobingsen',2,22,'beijing','xinandaxue');
insert into information values('luoluo',3,23,'shanghai','xinandaxue');
insert into information values('lbs',4,24,'tianjin','xinandaxue');

此时info 表中数据如下

SQL> select * from info;

NAME              AGE ADDRESS
---------- ---------- ----------
robinson          100 chongqing
luobingsen        100 chongqing
luoluo            100 chongqing

information 表中数据如下

SQL> select * from informATION;

NAME               ID        AGE ADDRESS    SCHOOL
---------- ---------- ---------- ---------- ----------
robinson            1         21 chongqing  xinandaxue
luobingsen          2         22 beijing    xinandaxue
luoluo              3         23 shanghai   xinandaxue
lbs                 4         24 tianjin    xinandaxue

现在，我想把information表name,age,address的信息插入到 info表中，

如果info 表中没有 information的信息，那么直接将该信息添加到info表

如果info 表中有information的信息，那么以information的信息为准，更新info表

SQL> merge into info i
  2  using (select name,age,address from information) j
  3  on (i.name=j.name)
  4  when matched then update set i.age=j.age,i.address=j.address
  5  when not  matched then insert values(j.name,j.age,j.address);

4 rows merged.

SQL> select * from info;

NAME              AGE ADDRESS
---------- ---------- ----------
robinson           21 chongqing
luobingsen         22 beijing
luoluo             23 shanghai
lbs                24 tianjin