Question

我不确定我的代码有什么问题。

 delimiter $$
 CREATE TRIGGER updateRestaurantAtributes 
 AFTER UPDATE ON fields_data
 FOR EACH ROW BEGIN
 IF (NEW.fieldid = 1) THEN
    UPDATE restaurants
    SET address1 = NEW.data_txt 
    Where rid = NEW.itemid;
 ELSE IF (NEW.fieldid = 2) THEN
    UPDATE restaurants
    SET address2 = NEW.data_txt 
    Where rid = NEW.itemid;
 END IF;
END$$

上述版本不起作用。它说“END”（最后一行）附近的语法错误。但是当我使用

时，某些东西会起作用

delimiter $$
CREATE TRIGGER updateRestaurantAtributes 
AFTER UPDATE ON fields_data
FOR EACH ROW BEGIN
IF (NEW.fieldid = 1) THEN
    UPDATE restaurants
    SET address1 = NEW.data_txt 
    Where rid = NEW.itemid;
END IF;
IF (NEW.fieldid = 2) THEN
    UPDATE restaurants
    SET address2 = NEW.data_txt 
    Where rid = NEW.itemid;
END IF;
END$$

我不确定为什么。我错过了什么吗？

Answer 1

而不是ELSE IF，MySQL's syntax使用ELSEIF（没有空格）。

 delimiter $$
 CREATE TRIGGER updateRestaurantAtributes 
 AFTER UPDATE ON fields_data
 FOR EACH ROW BEGIN
 IF (NEW.fieldid = 1) THEN
    UPDATE restaurants
    SET address1 = NEW.data_txt 
    Where rid = NEW.itemid;
 ELSEIF (NEW.fieldid = 2) THEN
    UPDATE restaurants
    SET address2 = NEW.data_txt 
    Where rid = NEW.itemid;
 END IF;
END$$

虽然您可以通过添加额外的ELSE IF来使其与END IF中的空格一起使用。通过使用空格，您可以有效地启动第二个IF语句，该语句必须独立于第一个外部IF语句而关闭。

/* Might work */
delimiter $$
 CREATE TRIGGER updateRestaurantAtributes 
 AFTER UPDATE ON fields_data
 FOR EACH ROW BEGIN
 IF (NEW.fieldid = 1) THEN
    UPDATE restaurants
    SET address1 = NEW.data_txt 
    Where rid = NEW.itemid;
 /* Opens a seconds IF block which must be closed */
 ELSE IF (NEW.fieldid = 2) THEN
    UPDATE restaurants
    SET address2 = NEW.data_txt 
    Where rid = NEW.itemid;
  /* Close inner IF block */
  END IF;
 END IF;
END$$

Answer 2

在Mysql Workbench 8.0中，单击表旁边的设置图标，然后在其中找到触发器选项卡。根据您的要求创建触发器，遵循if-else语法将完美地工作

CREATE DEFINER=`root`@`localhost` TRIGGER `users_AFTER_INSERT` AFTER INSERT ON `users` FOR EACH ROW BEGIN

 IF (NEW.status_active = 1) THEN
    UPDATE broadcast_status SET `type` = 0, `status` = 0 WHERE user_id = NEW.id;

 ELSE 
        IF (NEW.status_active = 2) THEN
            UPDATE broadcast_status SET `type` = 0, `status` = 2 WHERE user_id = NEW.id;
        ELSE 
            UPDATE broadcast_status SET `type` = 0, `status` = 1 WHERE user_id = NEW.id;
        END IF;
 END IF;

END

在sql触发器中使用else时出错

2 个答案: