Question

我写了触发器来检查客户是否有任何暂停帐户。如果是，他将无法创建新帐户，但是当我尝试插入时，插入仍然会通过。有人可以帮我这个吗？

CREATE OR REPLACE TRIGGER SuspendedAccount
BEFORE INSERT ON ACCOUNT

FOR EACH ROW

DECLARE

varIsProblemAccount Int;

BEGIN

SELECT COUNT(*) INTO varIsProblemAccount
FROM ACCOUNT
WHERE CUSTID = :new.custID
    AND ACCSTATUS = 'Suspended';

IF (varIsProblemAccount >= 1) THEN      
        DBMS_OUTPUT.PUT_LINE('*************************************************');
        DBMS_OUTPUT.PUT_LINE('');
        DBMS_OUTPUT.PUT_LINE('   Transaction canceled. ');
        DBMS_OUTPUT.PUT_LINE('');
        DBMS_OUTPUT.PUT_LINE('   Refer customer to the manager immediately. ');
        DBMS_OUTPUT.PUT_LINE('');
        DBMS_OUTPUT.PUT_LINE('**************************************************');
        RETURN;
    END;

ELSIF (varIsProblemAccount = 0) THEN

    BEGIN
        DBMS_OUTPUT.PUT_LINE('*************************************************');
        DBMS_OUTPUT.PUT_LINE('');
        DBMS_OUTPUT.PUT_LINE('   Transaction completed. ');
        DBMS_OUTPUT.PUT_LINE('');
        DBMS_OUTPUT.PUT_LINE('   Thank the customer for their business. ');
        DBMS_OUTPUT.PUT_LINE('');
        DBMS_OUTPUT.PUT_LINE('**************************************************');
    END;
END IF;
END;

抱歉，我试图更改为0，因为它永远不会是“NULL”但它仍然无法正常工作

Answer 1

问题是你的其他条件！

CREATE OR REPLACE TRIGGER SuspendedAccount
BEFORE INSERT ON ACCOUNT

FOR EACH ROW

DECLARE

varIsProblemAccount Int;

BEGIN

SELECT COUNT(*) INTO varIsProblemAccount
FROM ACCOUNT
WHERE CUSTID = :new.custID
    AND ACCSTATUS = 'Suspended';

IF (varIsProblemAccount >= 1) THEN      
        DBMS_OUTPUT.PUT_LINE('*************************************************');
        DBMS_OUTPUT.PUT_LINE('');
        DBMS_OUTPUT.PUT_LINE('   Transaction canceled. ');
        DBMS_OUTPUT.PUT_LINE('');
        DBMS_OUTPUT.PUT_LINE('   Refer customer to the manager immediately. ');
        DBMS_OUTPUT.PUT_LINE('');
        DBMS_OUTPUT.PUT_LINE('**************************************************');
        RETURN;
    END;

-- count() never returns NULL
ELSIF (varIsProblemAccount = 0) THEN

    BEGIN
        DBMS_OUTPUT.PUT_LINE('*************************************************');
        DBMS_OUTPUT.PUT_LINE('');
        DBMS_OUTPUT.PUT_LINE('   Transaction completed. ');
        DBMS_OUTPUT.PUT_LINE('');
        DBMS_OUTPUT.PUT_LINE('   Thank the customer for their business. ');
        DBMS_OUTPUT.PUT_LINE('');
        DBMS_OUTPUT.PUT_LINE('**************************************************');
    END;
END IF;
END;

Answer 2

尝试更改以下代码

ELSIF (varIsProblemAccount = null) THEN

到

ELSIF (varIsProblemAccount = 0) THEN

Answer 3

您的触发器无法阻止INSERT - RETURN过早退出触发器主体。

如果要阻止INSERT，则应该引发异常：

IF (varIsProblemAccount >= 1) THEN      
    raise_application_error(-20001, 'Existing account found');
END;

触发和计数（*）不起作用

3 个答案: