有TA, TB两表,假设均有三个栏位id, name, remark. 现在需要把TB表的name, remark两个栏位通过id关联,更新到TA表的对应栏位。
建表脚本:
- drop table TA;
- create table TA
- (
- id number not null,
- name varchar(10) not null,
- remark varchar(10) not null
- );
- drop table TB;
- create table TB
- (
- id number not null,
- name varchar(10) not null,
- remark varchar(10) not null
- );
- truncate table TA;
- insert into TA values(1, 'Aname1', 'Aremak1');
- insert into TA values(2, 'Aname2', 'Aremak2');
- commit;
- truncate table TB;
- insert into TB values(1, 'Bname1', 'Bremak1');
- insert into TB values(3, 'Bname3', 'Bremak3');
- commit;
- select * from TA;
- select * from TB;
SQLServer/Oracle版本的Update写法分别如下:
1. SQLServer
- update TA set name=b.name, remark=b.remark from TA a inner join TB b on a.id = b.id
或者
- update TA set name=b.name, remark=b.remark from TA a, TB b where a.id = b.id
注意不要在被更新表的的栏位前面加别名前缀,否则语法静态检查没问题,实际执行会报错。
Msg 4104, Level 16, State 1, Line 1
The multi-part identifier "a.name" could not be bound.
2. Oracle
- update TA a set(name, remark)=(select b.name, b.remark from TB b where b.id=a.id)
- where exists(select 1 from TB b where b.id=a.id)
注意如果不添加后面的exists语句,TA关联不到的行name, remark栏位将被更新为NULL值, 如果name, remark栏位不允许为null,则报错。 这不是我们希望看到的。
- --when name, remark is not null, cause error.
- --if allow null, rows in TA not matched will be update to null.
- update TA a set(name, remark)=(select b.name, b.remark from TB b where b.id=a.id);
可考虑的替代方法:
- update TA a set name= nvl((select b.name from TB b where b.id=a.id), a.name);
- update TA a set remark= nvl((select b.remark from TB b where b.id=a.id), a.remark);
如果TA.id, TB.id是unique index或primary key
可以使用视图更新的语法:
- ALTER TABLE TA ADD CONSTRAINT TA_PK
- PRIMARY KEY (
- ID
- )
- ENABLE
- VALIDATE
- ;
- ALTER TABLE TB ADD CONSTRAINT TB_PK
- PRIMARY KEY (
- ID
- )
- ENABLE
- VALIDATE
- ;
- update (select a.name, b.name as newname,
- a.remark, b.remark as newremark from TA a, TB b where a.id=b.id)
- set name=newname, remark=newremark;
-
-
- -------------------------------------------------------------
- Oracle有下面两个表:将表tab1中id值与和表tab2中id值相同的行的val更新为tab2中val的值.
select * from tab1;select * from tab2
更新完后的结果是:select * from tab1,在tab1中有的行,如果在tab2中没有对应的行,值被更新为null
where exists (select 1 from tab2 where tab1.id=tab2.id)
但是如果tab2中有多条对应tab1中一条的情况也会出错.
最好的方法是用merge语法:- merge into tab1
- using tab2
- on(tab1.id=tab2.id)
- when matched then
- update set tab1.val = tab2.val
比如在tab2中再插入一条 insert into tab2 values(2,'xxxx')
可以通过在using中的subquery中将重复记录过滤来避免这种错误,merge终极版:- merge into tab1
- using (select * FROM tab2 X WHERE X.ROWID =
- (SELECT MAX(Y.ROWID) FROM tab2 Y WHERE X.ID = Y.ID)) tab2
- on(tab1.id=tab2.id)
- when matched then
- update set tab1.val = tab2.val