Question

我尝试使用触发器在我插入的新值未遵循先前值时显示RAISE_APPLICATION_ERROR。例如，我的最新位置在第二位置，当我想插入新位置10时，它应显示ORA-…“我输入的错误消息”？任何人都可以帮助我编辑我的代码？我需要一个选择语句吗？

SET ECHO ON
SET FEEDBACK ON
SET serveroutput on size 2000 

CREATE OR REPLACE TRIGGER RowTrigger
Before 
 INSERT OR UPDATE on POSITION
 FOR EACH ROW
 declare
 u number(2);

begin
 select count(P#) into u from POSITION ;

  if :new.P# < u
  then
 RAISE_APPLICATION_ERROR(-20011,'dddd');
 end if;

 end;

Answer 1

好的。有时候，我花了一些时间来编写它，下面的工作考虑了这个问题：

-The P# is unique, or youll get error in the select.
-I am not taking on consideration the previous data , for example  if your P# is 5 and ur inserting 3, then youll not have error.
   if you need such thing you can continue developing it because there are some conditions to be met.
-The table already contains data , all you can add a exception NO_data_found and do whatever you want in that case.

这是代码：

drop table ex_employee
/
create table ex_employee (id number(2) null,id1 number(2) NULL,name varchar2(1000) null)

/
drop table POSITION
/
create table POSITION(p# number(2) null)
/
insert into POSITION(p#) values (4)
/
insert into POSITION(p#) values (5)
/
commit
/
CREATE OR REPLACE TRIGGER RowTrigger
  FOR INSERT on POSITION compound trigger
     prevID number(4);
     new_p number(4);
  after each row is
begin

  new_p := :new.p#;
end after each row;

  after statement is
     begin 
     --it gives the previous record;
        select PREV_ID into prevID from(SELECt p#, LAG(p#, 1, 0) OVER(ORDER BY p#) as PREV_ID FROM position) where p# = new_p;
      -- check if the old data is smaller then the new, you can add more conditions.
            if (previd < new_p) then
      previd:=previd+1;
      end if;

     IF (prevID != new_p) THEN 
     RAISE_APPLICATION_ERROR(-20011, 'ERROR , data not in order or whatever.');
     END IF; 


   end  after statement;

end;
/

当我插入的值不符合先前的值时显示RAISE_APPLICATION_ERROR

1 个答案: