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update关联更新在sqlserver和oracle中的实现

原创 Oracle 作者:yaanzy 时间:2006-10-11 13:21:27 0 删除 编辑

sqlserver和oracle中实现update关联更新的语法不同,都可以通过inline view(内嵌视图)
来实现,总的来说sqlserver更简单些. 测试例子如下:

[@more@]

create table tmp_a
(cpcode varchar2(10),
sb_ym varchar2(6),
flag char(1)
);

create table tmp_b
(cpcode varchar2(10),
sb_ym varchar2(6),
flag char(1)
);

insert into tmp_a(cpcode,sb_ym,flag)values('3201910001','200406','e');
insert into tmp_a(cpcode,sb_ym,flag)values('3201910002','200406','e');
insert into tmp_b(cpcode,sb_ym,flag)values('3201910001','200406','r');
insert into tmp_b(cpcode,sb_ym,flag)values('3201910002','200406','r');
insert into tmp_b(cpcode,sb_ym,flag)values('3201910003','200406','r');
insert into tmp_b(cpcode,sb_ym,flag)values('3201910004','200406','e');
commit;

在SQLSERVER中:

update tmp_b set flag = b.flang from tmp_a a,tmp_b b
where a.cpcode =b.cpcode and a.sb_ym = b.sb_ym;


在Oracle中:

方法一:(效率低)
update tmp_b a
set flag = (select flag from tmp_a b
where a.cpcode = b.cpcode and a.sb_ym = b.sb_ym )
where exists
(select * from tmp_a c
where a.cpcode = c.cpcode and a.sb_ym = c.sb_ym);

Statistics
----------------------------------------------------------
8 recursive calls
3 db block gets
18 consistent gets
0 physical reads
0 redo size

方法二:(效率高)
alter table tmp_a add constraint p_tmp_a primary key (cpcode, sb_ym);

update (select b.flag flagb,a.flag flaga
from tmp_a a,tmp_b b
where a.cpcode=b.cpcode
and a.sb_ym=b.sb_ym)
set flagb=flaga;

Statistics
----------------------------------------------------------
0 recursive calls
3 db block gets
7 consistent gets
0 physical reads
0 redo size


注意:方法二中数据源表必须要加上主键,否则会报错
ORA-01779: 无法修改与非键值保存表对应的列
被修改的表则无需增加主键

来自 “ ITPUB博客 ” ,链接:http://blog.itpub.net/3898/viewspace-871738/,如需转载,请注明出处,否则将追究法律责任。

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