如何在插入MySQL之前尝试创建触发器时抛出错误,以便为每个输入生成新的UUID.
CREATE TRIGGER insert_guid BEFORE INSERT ON guid_tool (FOR EACH ROW BEGIN SET NEW.guid_key = UUID()); END;
这是我的桌子
create table guid_tool ( ID INT NOT NULL AUTO_INCREMENT, guid_key CHAR(40) NOT NULL, PRIMARY KEY(ID) ) CHARSET=LATIN1;
我一定做错了什么.
正如@mabi在评论中所提到的 - 括号中存在语法错误.以下修改后的代码适用于我:
DELIMITER # CREATE TRIGGER insert_guid BEFORE INSERT ON guid_tool FOR EACH ROW BEGIN SET NEW.guid_key = UUID(); END; # DELIMITER ;
测试出来:
mysql> INSERT INTO guid_tool (ID) VALUES (1); Query OK, 1 row affected, 1 warning (0.04 sec) mysql> SELECT * FROM guid_tool; +----+--------------------------------------+ | ID | guid_key | +----+--------------------------------------+ | 1 | a0467ebf-5c4f-11e3-903a-6cccbb4423e3 | +----+--------------------------------------+ 1 row in set (0.00 sec)