with r as {
select CLM001 ,sum(score) as total from b group by CLM001
}
update a set score = (select total from r where r.CLM001 =a.CLM001)
where exists (select 1 from r where r.CLM001 =a.CLM001)
点评:此方法需要写存储过程，否则在程序中没法用。
create or replace procedure p_test
as
v_clm001 a.clm001%type;
cursor c_record is
select distinct clm001 from b;
begin
open c_record;
loop
fetch c_record into v_clm001;
exit when c_record%notfound;
update a set score =(select sum(score) from b where clm001=v_clm001)
where clm001=v_clm001;
end loop;
end;
点评:这是在PL/SQL中的语句，此问题用PL/SQL解决也不失为一个简单的方法。
update a set a.score = a.score + (select b.score from a a,
(select CLM001,sum(score) from b group by CLM001) b where a.CLM001 = b.CLM001)
点评:最佳答案，此类问题用视图解决很方便!
