今天写了个复杂的SQL,用来更新另一个表

select vin,(max(in_mileage)-min(in_mileage))/(max(start_time)-min(start_time)) from (
 select vin,in_mileage,start_time
 from (select vin,in_mileage,start_time ,
 row_number() over (partition by vin order by start_time) wwmnum from
(select  vin,max(in_mileage) in_mileage,max(start_time) start_time from
(select vin,in_mileage,start_time from tt_repair_order
  union
select vin,in_mileage,start_time  from tt_ro_balanced)
  group by vin,to_char(start_time,'yyyymmdd')
)
)
 where wwmnum<3
 )  group by vin having max(start_time)-min(start_time)<>0 and to_char(max(start_time),'yyyymmdd')<>to_char(min(start_time),'yyyymmdd')

写完这个SQL交给程序员后,程序员跟我说不会UPDATE,也就是说通过B表更新A表程序员是有困难的,于是在这里整理下各种不同的方法以供以后使用.

TOWN                         ID
-------------------- ----------
222                         222
111                         111
ww'jj                       111
llll                       1111
dddd                       2222
lllldf                      111
lllldf                      111
dsafdf                      111
3435                        111
ljjjjj                      222
dsafdf                      111

TOWN                         ID
-------------------- ----------
3435                        111
ljjjjj                      222

SQL> select * from wwm5;            --更新的条件表

TOWN                         ID
-------------------- ----------
lllldf                      111
test                       9984

SQL> select wwm2.* from wwm2,wwm5 where wwm2.id=wwm5.id
  2  /

TOWN                         ID
-------------------- ----------
111                         111
ww'jj                       111
lllldf                      111
lllldf                      111
dsafdf                      111
3435                        111
dsafdf                      111
3435                        111

8 rows selected.

所以,每次需要更新8条数据就是正确的.

相信程序员是通过以下类似的SQL更新的,这是错误的,因为没有加WHERE
SQL>  update wwm2 set wwm2.town=(select wwm5.town from wwm5 where wwm5.id=wwm2.id)
  2  /

13 rows updated.

SQL> select * from wwm2;

TOWN                         ID
-------------------- ----------
                            222
lllldf                      111
lllldf                      111
                           1111
                           2222
lllldf                      111
lllldf                      111
lllldf                      111
lllldf                      111
                            222
lllldf                      111

TOWN                         ID
-------------------- ----------
lllldf                      111
                            222
13 rows selected.

可以看到13条记录被更新,符合条件的更新正确,不符合条件的也更新为NULL.以下是正确的方法

方法一:
SQL> update wwm2
  2  set town=(select town from wwm5 where wwm5.id=wwm2.id)
  3  where id=(select wwm5.id from wwm5 where wwm5.id=wwm2.id)
  4  /

8 rows updated.

方法二:    与方法一道理相同,这里需要掌握EXIST的相关用法.
SQL> update wwm2
   set town=(select town from wwm5 where wwm5.id=wwm2.id)
   where exists (select 1 from wwm5 where wwm5.id=wwm2.id)
8 rows updated.

方法三:
SQL> update (select a.town atown,a.id aid,b.town btown,b.id bid from wwm2 a,wwm5 b where a.id=b.id)
  2  set atown=btown
  3  /
set atown=btown
    *
ERROR at line 2:
ORA-01779: cannot modify a column which maps to a non key-preserved table

  1* alter table wwm5 add primary key (id)
SQL> /

Table altered.

  1  update (select a.town atown,a.id aid,b.town btown,b.id bid from wwm2 a,wwm5 b where a.id=b.id)
  2*  set atown=btown
SQL> /

8 rows updated.

这种方法的局限性就是需要PRIMARY 的支持.

方法四:
  1  declare
  2  cursor cur_wwm is select town,id from wwm5;
  3  begin
  4     for my_wwm in cur_wwm loop
  5     update wwm2 set town=my_wwm.town
  6     where id=my_wwm.id;
  7     end loop;
  8* end;
SQL> /

PL/SQL procedure successfully completed.

SQL> select * from wwm2;

TOWN                         ID
-------------------- ----------
222                         222
lllldf                      111
lllldf                      111
llll                       1111
dddd                       2222
lllldf                      111
lllldf                      111
lllldf                      111
lllldf                      111
ljjjjj                      222
lllldf                      111

TOWN                         ID
-------------------- ----------
lllldf                      111
ljjjjj                      222

这个方法是最灵活的了.

方法五:

注意,方法五只能适用于WWM5是WWM2的子集的时候.
  1   merge into wwm2
  2   using (select town,id from wwm5) b
  3   on (wwm2.id=b.id)
  4   when matched then update set town=b.town
  5* when not matched then insert (town,id) values (null,null)
SQL> /

9 rows merged.

SQL> select * from wwm2;

TOWN                         ID
-------------------- ----------
                                  ---注意这个地方,被插入了一个空值.因为WWM5的ID=9984在WWM2中不能匹配,根本原因是ORACLE9必须有WHEN NOT MATCHED子句,但是ORACLE10可以不许要,也就是ORACLE10可以不写WHEN NOT MATCHED ,就不必插入NULL值了,为解决这个问题,下一步会DELETE WWM5的ID=9984,这样一来就不会执行WHEN NOT MATCHED
222                         222
lllldf                      111
lllldf                      111
llll                       1111
dddd                       2222
lllldf                      111
lllldf                      111
lllldf                      111
lllldf                      111
ljjjjj                      222

TOWN                         ID
-------------------- ----------
lllldf                      111
lllldf                      111
ljjjjj                      222

14 rows selected.

SQL> delete from wwm5 where id=9984;

1 row deleted.

SQL>  1   merge into wwm2                            
SQL>   2   using (select town,id from wwm5) b
SQL>   3   on (wwm2.id=b.id)
SQL>   4   when matched then update set town=b.town
SQL>   5* when not matched then insert (town,id) values (null,null)
SQL> /

8 rows merged.

 

       以上就是5种关连更新的例子了,希望能给开发人员解惑.