转ORACLE批量更新四种方法比较

软件环境 Windows 2000 + ORACLE9i
硬件环境 cpu 1.8G + RAM 512M

现在我们有2张表 如下：
T1–大表 10000笔 T1_FK_ID
T2–小表 5000笔 T2_PK_ID
T1通过表中字段ID与T2的主键ID关联

模拟数据如下：
–T2有5000笔数据
create table T2
as
select rownum id,a.*
from all_objects a
where 1=0;

– Create/Recreate primary,unique and foreign key constraints
alter table T2
add constraint T2_PK_ID primary key (ID);

insert /+ APPEND / into T2
select rownum id,a.*
from all_objects a where rownum<=5000;

–T1有10000笔数据
create table T1
as
select rownum sid,T2.*
from T2
where 1=0;

– Create/Recreate primary,unique and foreign key constraints
alter table T1
add constraint T1_FK_ID foreign key (ID)
references t2 (ID);

insert /+ APPEND / into T1
select rownum sid,T2.*
from T2;

insert /+ APPEND / into T1
select rownum sid,T2.*
from T2;

–更新Subobject_Name字段，之前为null
update T2 set T2.Subobject_Name=’StevenHuang’

我们希望能把T1的Subobject_Name字段也全部更新成’StevenHuang’，也就是说T1的10000笔数据都会得到更新

方法一
写PL/sql,开cursor
declare
l_varID varchar2(20);
l_varSubName varchar2(30);
cursor mycur is select T2.Id,T2.Subobject_Name from T2;

begin
open mycur;
loop
fetch mycur into l_varID,l_varSubName;
exit when mycur %notfound;
update T1 set T1.Subobject_Name = l_varSubName where T1.ID = l_varID;
end loop;
close mycur;
end;
—耗时39.716s
显然这是最传统的方法，如果数据量巨大的话（4000万笔），还会报”snapshot too old”错误退出

方法二.
用loop循环，分批操作
declare
i number;
j number;
begin
i := 1;
j := 0;
select count(*) into j from T1;
loop
exit when i > j;
update T1 set T1.Subobject_Name = (select T2.Subobject_Name from T2 where T1.ID = T2.ID)
where T1.ID >= i and T1.ID <= (i + 1000);
i := i + 1000;
end loop;
end;
–耗时0.656s，这里一共循环了10次，如果数据量巨大的话，虽然能够完成任务，但是速度还是不能令人满意。(例如我们将T1–大表增大到100000笔 T2–小表增大到50000笔
) 耗时10.139s

方法三.
–虚拟一张表来进行操作，在数据量大的情况下效率比方法二高很多
update (select T1.Subobject_Name a1,T2.Subobject_Name b1 from T1,T2 where T1.ID=T2.ID)
set a1=b1;
–耗时3.234s (T1–大表增大到100000笔 T2–小表增大到50000笔)

方法四.
–由于UPDATE是比较消耗资源的操作，会有redo和undo操作，在这个例子里面我们可以换用下面的方法，创建一张新表，因为采用insert比update快的多，之后你会有一张旧表和一张新表，然后要怎么做就具体情况具体分析了~
create table T3 as select * from T1 where rownum<1;
alter table T3 nologging;
insert /+ APPEND / into T3
select T1.* from T1,T2 where T1.ID=T2.ID;
–耗时0.398s (T1–大表增大到100000笔 T2–小表增大到50000笔)

*以上所有操作都已经将分析执行计划所需的时间排除在外