抱歉我的英文。
我有两张桌子:
Table1
id
table2_id
num
modification_date
和
Table2
id
table2num
我希望在Table1
中插入或删除后更新num
中的最后一个值Table2.table1lastnum
。
我的触发器:
CREATE OR REPLACE TRIGGER TABLE1_NUM_TRG
AFTER INSERT OR DELETE ON table1
FOR EACH ROW
BEGIN
IF INSERTING then
UPDATE table2
SET table2num = :new.num
WHERE table2.id = :new.table2_id;
ELSE
UPDATE table2
SET table2num = (SELECT num FROM (SELECT num FROM table1 WHERE table2_id = :old.table2_id ORDER BY modification_date DESC) WHERE ROWNUM <= 1)
WHERE table2.id = :old.table2_id;
END IF;
END TABLE1_NUM_TRG;
但是在Table1
删除后我有错误:
ORA-04091: table BD.TABLE1 is mutating, trigger/function may not see it
ORA-06512: at "BD.TABLE1_NUM_TRG", line 11
ORA-04088: error during execution of trigger 'BD.TABLE1_NUM_TRG'
我做错了什么?
答案 0 :(得分:6)
您遇到的是经典的“变异表”异常。在ROW触发器中,Oracle不允许您对定义触发器的表运行查询 - 因此它是引发此问题的触发器SELECT
部分中针对TABLE1的DELETING
。 / p>
有几种方法可以解决这个问题。在这种情况下,最好的方法是使用复合触发器,它看起来像:
CREATE OR REPLACE TRIGGER TABLE1_NUM_TRG
FOR INSERT OR DELETE ON TABLE1
COMPOUND TRIGGER
TYPE NUMBER_TABLE IS TABLE OF NUMBER;
tblTABLE2_IDS NUMBER_TABLE;
BEFORE STATEMENT IS
BEGIN
tblTABLE2_IDS := NUMBER_TABLE();
END BEFORE STATEMENT;
AFTER EACH ROW IS
BEGIN
IF INSERTING THEN
UPDATE TABLE2 t2
SET t2.TABLE2NUM = :new.NUM
WHERE t2.ID = :new.TABLE2_ID;
ELSIF DELETING THEN
tblTABLE2_IDS.EXTEND;
tblTABLE2_IDS(tblTABLE2_IDS.LAST) := :new.TABLE2_ID;
END IF;
END AFTER EACH ROW;
AFTER STATEMENT IS
BEGIN
IF tblTABLE2_IDS.COUNT > 0 THEN
FOR i IN tblTABLE2_IDS.FIRST..tblTABLE2_IDS.LAST LOOP
UPDATE TABLE2 t2
SET t2.TABLE2NUM = (SELECT NUM
FROM (SELECT t1.NUM
FROM TABLE1 t1
WHERE t1.TABLE2_ID = tblTABLE2_IDS(i)
ORDER BY modification_date DESC)
WHERE ROWNUM = 1)
WHERE t2.ID = tblTABLE2_IDS(i);
END LOOP;
END IF;
END AFTER STATEMENT;
END TABLE1_NUM_TRG;
复合触发器允许处理每个时间点(BEFORE STATEMENT
,BEFORE ROW
,AFTER ROW
和AFTER STATEMENT
)。请注意,始终按给定的顺序调用时间点。当执行适当的SQL语句(即INSERT INTO TABLE1
或DELETE FROM TABLE1
)并触发此触发器时,要调用的第一个时间点将为BEFORE STATEMENT
,并且{{1}中的代码处理程序将分配一个PL / SQL表来保存一堆数字。在这种情况下,要存储在PL / SQL表中的数字将是TABLE1中的TABLE2_ID值。 (例如,使用PL / SQL表而不是数组,因为表可以容纳不同数量的值,而如果我们使用数组,我们必须事先知道需要存储多少个数。我们事先不能知道特定语句会影响多少行,因此我们使用PL / SQL表)。当达到BEFORE STATEMENT
时间点并且我们发现正在处理的语句是INSERT时,触发器就会继续执行必要的UPDATE到TABLE2,因为这不会导致问题。但是,如果正在执行DELETE,则触发器将TABLE1.TABLE2_ID保存到先前分配的PL / SQL表中。当最终到达AFTER EACH ROW
时间点时,先前分配的PL / SQL表被迭代,并且对于找到的每个TABLE2_ID,执行适当的更新。
分享并享受。
答案 1 :(得分:1)
你必须为delete定义一个before触发器。尝试使用两个触发器
CREATE OR REPLACE TRIGGER INS_TABLE1_NUM_TRG
AFTER INSERT ON table1
FOR EACH ROW
BEGIN
UPDATE table2
SET table2num = :new.num
WHERE table2.id = :new.table2_id;
END INS_TABLE1_NUM_TRG;
CREATE OR REPLACE TRIGGER DEL_TABLE1_NUM_TRG
BEFORE DELETE ON table1
FOR EACH ROW
BEGIN
UPDATE table2
SET table2num = (SELECT num FROM
(SELECT num FROM table1 WHERE table2_id = :old.table2_id
ORDER BY modification_date DESC)
WHERE ROWNUM <= 1)
WHERE table2.id = :old.table2_id;
END DEL_TABLE1_NUM_TRG;
答案 2 :(得分:0)
@psaraj12 答案是最好的恕我直言,但在 DELETE 触发器中,我会使用 :OLD 表示法,因为内部查询是不必要的,并且会显着减慢触发速度:
...
BEFORE DELETE ON table1
FOR EACH ROW
UPDATE table2
SET table2num = :OLD.num
WHERE table2.id = :OLD.table2_id;
...