Question

我遇到与在phpmyadmin上创建触发器有关的问题。我有一个简单的代码，如下所示，当我提交代码时，在变量x声明的行中返回一个sintax错误。我知道SELECT查询有效，因为我测试了它，并返回一个字符串。谁能帮我？提前谢谢。

DELIMITER $$

CREATE TRIGGER `after_assigntags_insert` AFTER INSERT ON `assigntags`
FOR EACH ROW 
BEGIN
DECLARE x VARCHAR(10);
SET x = (SELECT UltLabel FROM assignlabels WHERE ALmessage=1);
INSERT INTO messages (Message) VALUES (x);
$$
END

DELIMITER ;

Answer 1

如何使用user variables？

DELIMITER $$

CREATE TRIGGER `after_assigntags_insert` AFTER INSERT ON `assigntags`
FOR EACH ROW 
BEGIN
   SET @x = (SELECT UltLabel FROM assignlabels WHERE ALmessage=1 LIMIT 1);
   INSERT INTO messages (Message) VALUES (@x);
$$
END

DELIMITER ;

Answer 2

我认为BEGIN来自DECLARE

DELIMITER $$

CREATE TRIGGER `after_assigntags_insert` AFTER INSERT ON `assigntags`
FOR EACH ROW 

DECLARE 
  x VARCHAR(10);
BEGIN
  SET x = (SELECT UltLabel FROM assignlabels WHERE ALmessage=1);
  INSERT INTO messages (Message) VALUES (x);
  $$
END

DELIMITER ;

Answer 3

这是语法错误，

END应该在您的分隔符$$之前。

尝试这样的事情：

DELIMITER $$

CREATE TRIGGER `after_assigntags_insert` AFTER INSERT ON `assigntags`
FOR EACH ROW 
BEGIN
DECLARE x VARCHAR(10);
SET x = (SELECT UltLabel FROM assignlabels WHERE ALmessage=1);
INSERT INTO messages (Message) VALUES (x);
END
$$

DELIMITER ;

触发器创建无效

3 个答案: