我想创建一个触发器,但在某处出现错误,我不知道如何编写正确的。Mysql创建触发器
CREATE TABLE IF NOT EXISTS Authorization(
userID INT PRIMARY KEY AUTO_INCREMENT,
login VARCHAR(50) NOT NULL,
password VARCHAR(64) NOT NULL,
numberOfFailedAttempts SMALLINT,
lastLogin TIMESTAMP,
lockedAccount BOOLEAN
);
INSERT INTO Authorization SELECT NULL, "test1", SHA2('pass1', 256), 1, NULL, FALSE;
INSERT INTO Authorization SELECT NULL, "test2", SHA2('pass2', 256), 1, NULL, FALSE;
INSERT INTO Authorization SELECT NULL, "test3", SHA2('pass3', 256), 1, NULL, FALSE;
INSERT INTO Authorization SELECT NULL, "test4", SHA2('pass4', 256), 1, NULL, FALSE;</code>
我想创建一个触发器,当更新用户检查numberOfFailedAttempts
。如果numberOfFailedAttempts
> 3,则将lockedAccount
更改为true。
CREATE TRIGGER trigger1 AFTER UPDATE ON `numberOfFailedAttempts`
FOR EACH ROW
BEGIN
IF (`numberOfFailedAttempts` > 3) THEN
UPDATE SET `lockedAccount` = 1 WHERE 1, LIMIT 1;
END;
的错误是:
#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'SET `lockedAccount` = 1 WHERE 1, LIMIT 1' at line 5</code>
有人可以帮我吗?
编辑:
在这段代码中我有同样的错误。
CREATE TRIGGER trigger1 AFTER UPDATE ON `numberOfFailedAttempts`
FOR EACH ROW
BEGIN
IF (`numberOfFailedAttempts` > 3) THEN
UPDATE `Authorization` SET `lockedAccount` = '1' WHERE `Authorization`.`userID` = OLD.`userID`;
END;
编辑:
我改变代码和它的工作。
delimiter //
CREATE TRIGGER trigger1 BEFORE UPDATE ON Authorization
FOR EACH ROW
IF NEW.`numberOfFailedAttempts` > 3 THEN
SET NEW.`lockedAccount` = 1;
END IF;//
错误表明,你需要UPDATE'语句后'指定表名 –
感谢您的答复,但在代码中仍然有同样的错误。 CREATE TRIGGER TRIGGER1 UPDATE ON Authorization.numberOfFailedAttempts FOR EACH ROW 之后开始 IF('numberOfFailedAttempts'> 3)THEN UPDATE SET OLD.'lockedAccount' = 1 WHERE 1,LIMIT 1; END; – pawel112
您在错误的地方设置了表名,试试这个:UPDATE授权SET lockedAccount = 1 WHERE 1,LIMIT 1;' –