Question

我在oracle中创建了一个触发器：

/* Formatted on 3-Oct-2013 15:58:45 (QP5 v5.126) */
CREATE OR REPLACE TRIGGER testtrigger
AFTER INSERT OR UPDATE OF sellpoint_name
ON sell_point
FOR EACH ROW
WHEN (new.sellpoint_name = 'Location')
DECLARE lat, lng sell_point.sellpoint_lat%TYPE;
BEGIN
SELECT  sellpoint_lat, sellpoint_long into lat, lng
  FROM  sell_point
 WHERE  sellpoint_name = :new.sellpoint_name;

IF (:new.sellpoint_lat < 20 OR :new.sellpoint_long < 100)
THEN
    raise_application_error (-20225, 'this point is not exists');
END IF;
END;

但是我收到了错误：

1/12    PLS-00103: Encountered the symbol "," when expecting one of the following:
constant exception <an identifier>
<a double-quoted delimited-identifier> table long double ref
char time timestamp interval date binary national character
nchar
1/47    PLS-00103: Encountered the symbol ";" when expecting one of the following:
16:10:50            := ( , not null range default external character

这里有什么不对？谢谢你的帮助！

Answer 1

试试这个，

CREATE OR REPLACE TRIGGER testtrigger
AFTER INSERT OR UPDATE OF sellpoint_name ON sell_point
FOR EACH ROW
WHEN (new.sellpoint_name = 'Location')
DECLARE 
     lat sell_point.sellpoint_lat%TYPE; 
     lng sell_point.sellpoint_long%TYPE;
BEGIN
     SELECT sellpoint_lat, sellpoint_long 
       INTO lat, lng
       FROM sell_point
      WHERE sellpoint_name = :new.sellpoint_name;

     IF (:NEW.sellpoint_lat < 20 OR :NEW.sellpoint_long < 100) THEN
          raise_application_error (-20225, 'this point is not exists');
     END IF;
END;

<强>被修改

当您尝试从同一个表中选择列时，上面的代码会引发错误table is mutating。

您可以编辑这样的代码，

CREATE OR REPLACE TRIGGER testtrigger
AFTER INSERT OR UPDATE OF sellpoint_name ON sell_point
FOR EACH ROW
WHEN (new.sellpoint_name = 'Location')
BEGIN
     IF (:NEW.sellpoint_lat < 20 OR :NEW.sellpoint_long < 100) THEN
          raise_application_error (-20225, 'this point is not exists');
     END IF;
END;

Answer 2

我认为因为你没有lat的数据类型

DECLARE lat, lng sell_point.sellpoint_lat%TYPE;
           ^

SQL创建触发器oracle时出错

2 个答案: