Question

我有某种继承的情况：

设备-> raut_controller-> raut_flat_temperature_control_controller

我创建了触发器来插入父行，以便在插入继承的表之前可以先获得其ID。在插入“ raut_controller”之前，触发器在device表中插入行，使用LAST_INSERT_ID（）获取ID并更改NEW.id。与raut_flat_temperature_control_controller相同。因此，在raut_flat_temperature_control_controller中插入应该在3个表中调用经修改的插入。问题是由于失败了外键约束，我无法在表raut_flat_temperature_control_controller中插入。毕竟触发它尝试插入到id = 1的“ raut_flat_temperature_control_controller”中。在父表“ raut_controller”中不存在。在raut_controller中触发时，已成功在device表中插入行。

如果raut_flat_temperature_control_controller为何LAST_INSERT_ID（）在触发器中不起作用？

UPD。有什么办法可以忽略mysql中的BEFORE INSERT TRIGGER

我的桌子：

CREATE TABLE `device` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `device_type_id` int(11) NOT NULL,
  PRIMARY KEY (`id`),
  KEY `idx_device_type_id` (`device_type_id`),
  CONSTRAINT `fk_device_type_id` FOREIGN KEY (`device_type_id`) REFERENCES `device_type` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION
);

CREATE TABLE `raut_controller` (
  `id` int(11) NOT NULL,
  `raut_controller_type_id` int(11) NOT NULL,
  `ip_adress` int(10) unsigned NOT NULL DEFAULT '0',
  `mac_address` bigint(20) unsigned NOT NULL DEFAULT '0',
  `serial_num` bigint(20) unsigned NOT NULL DEFAULT '0',
  PRIMARY KEY (`id`),
  KEY `fk_raut_controller_type_id_idx` (`raut_controller_type_id`),
  CONSTRAINT `fk_raut_controller_device` FOREIGN KEY (`id`) REFERENCES `device` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION,
  CONSTRAINT `fk_raut_controller_type_id` FOREIGN KEY (`raut_controller_type_id`) REFERENCES `raut_controller_type` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION
);

CREATE TABLE `raut_flat_temperature_control_controller` (
  `id` int(11) NOT NULL,
  `version` int(11) unsigned DEFAULT NULL,
  PRIMARY KEY (`id`),
  CONSTRAINT `fk_raut_flat_temperature_control` FOREIGN KEY (`id`) REFERENCES `raut_controller` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION
);

和2个查询表以获取* type_id：

CREATE TABLE `device_type` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(255) NOT NULL,
  `table_name` varchar(255) NOT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `unq_device_type_name` (`name`),
  UNIQUE KEY `unq_device_type_table_name` (`table_name`),
  KEY `idx_device_type_name` (`name`),
  KEY `idx_device_type_table_name` (`table_name`)
);

CREATE TABLE `raut_controller_type` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(255) NOT NULL,
  `table_name` varchar(255) NOT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `unq_raut_controller_type_name` (`name`),
  UNIQUE KEY `unq_raut_controller_type_table_name` (`table_name`),
  KEY `idx_raut_controller_type_name` (`name`),
  KEY `id_raut_controller_type_table_name` (`table_name`)
);

触发器

CREATE TRIGGER `raut_controller_device_id` BEFORE INSERT ON `raut_controller` FOR EACH ROW
BEGIN
        INSERT INTO `reeve`.`device` (
            `device_type_id`
        )
        VALUES (
            (SELECT `id` FROM `reeve`.`device_type` WHERE table_name = 'raut_controller')
        );
        SET NEW.id = LAST_INSERT_ID();
END

CREATE TRIGGER `raut_flat_temperature_control_controller_id` BEFORE INSERT ON `raut_flat_temperature_control_controller` FOR EACH ROW
BEGIN
        INSERT INTO `reeve`.`raut_controller` (
            `raut_controller_type_id`
        )
        VALUES (
            (SELECT `id` FROM `reeve`.`raut_controller_type` WHERE table_name = 'raut_flat_temperature_control_controller')
        );
        SET NEW.id = LAST_INSERT_ID();        
END

LAST_INSERT_ID（）在触发器中始终为1

0 个答案: