Question

正在努力实现-

我正在尝试更新tb_sites_3中的color_status（3将根据从tb_tickets获得的program_id是动态的），只要在tb_jobs上进行任何插入。

错误

在创建触发器时，出现以下错误 错误＃1054-“新建”中的未知列“ program_id”

tb_tickets

tb_jobs

tb_sites_3

DELIMITER //
    CREATE TRIGGER trig_job_color
           BEFORE INSERT ON `tb_jobs`
           FOR EACH ROW 
    BEGIN
    SET NEW.program_id = (Select program_id from tb_tickets
    where tb_tickets.job_id = NEW.job_id);
    SET NEW.status = (Select status from tb_tickets
    where tb_tickets.job_id = NEW.job_id);

     CASE NEW.program_id
     WHEN 1 THEN
       UPDATE tb_sites_1 
       SET color_status = NEW.status 
       WHERE site_id = NEW.site_id;
     WHEN 2 THEN
       UPDATE tb_sites_2 
       SET color_status = NEW.status 
       WHERE site_id = NEW.site_id;
     WHEN 3 THEN
       UPDATE tb_sites_3
       SET color_status = NEW.status 
       WHERE site_id = NEW.site_id;
     END CASE;
    END //
    DELIMITER ;

表格定义

tb_tickets

CREATE TABLE `tb_tickets` (
 `id` int(15) NOT NULL,
 `ticket_id` int(15) NOT NULL,
 `job_id` int(11) NOT NULL,
 `site_id` varchar(200) NOT NULL,
 `program_id` int(11) NOT NULL,
 `status` varchar(200) NOT NULL,
 PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1

tb_jobs

CREATE TABLE `tb_jobs` (
 `job_id` int(11) NOT NULL AUTO_INCREMENT,
 `job_creation` date DEFAULT NULL,
 PRIMARY KEY (`job_id`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=latin1

tb_sites_3

CREATE TABLE `tb_sites_3` (
 `id` int(15) NOT NULL AUTO_INCREMENT,
 `color_status` int(15) NOT NULL,
 `site_id` varchar(200) NOT NULL,
 PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=latin1

Answer 1

在插入trig_job_color之后创建触发器tb_jobs 对于每个行开始 SET @program_id =（从tb_tickets中选择program_id 其中tb_tickets.job_id = NEW.job_id）;

SET @newstatus = (Select status from tb_tickets
where tb_tickets.job_id = NEW.job_id);
SET @newsite_id = (Select site_id from tb_tickets
where tb_tickets.job_id = NEW.job_id);

CASE @program_id
 WHEN 1 THEN
   UPDATE tb_sites_3 
   SET tb_sites_3.color_status = @newstatus 
   WHERE tb_sites_3.site_id = @newsite_id;
 WHEN 2 THEN
   UPDATE tb_sites_3 
   SET tb_sites_3.color_status = @newstatus 
   WHERE tb_sites_3.site_id = @newsite_id;
 WHEN 3 THEN
   UPDATE tb_sites_3
   SET tb_sites_3.color_status = @newstatus 
   WHERE tb_sites_3.site_id = @newsite_id;
 END CASE;
 END

Answer 2

这对我来说使用变量很有效。

DELIMITER //
    CREATE TRIGGER trig_job_color
           AFTER INSERT ON `tb_jobs`
           FOR EACH ROW 
    BEGIN
    DECLARE x, y INT DEFAULT 0;
    DECLARE z varchar(50);
    SET x = (Select program_id from tb_tickets
    where tb_tickets.job_id = NEW.job_id);
    SET y = (Select status from tb_tickets
    where tb_tickets.job_id = NEW.job_id);
    SET Z = (Select site_id from tb_tickets
    where tb_tickets.job_id = NEW.job_id);
     CASE x
     WHEN 1 THEN
       UPDATE tb_sites_1 
       SET color_status = y
       WHERE site_id = z;
     WHEN 2 THEN
       UPDATE tb_sites_2 
       SET color_status = y
       WHERE site_id = z;
     WHEN 3 THEN
       UPDATE tb_sites_3
       SET color_status = y 
       WHERE site_id = z;
     END CASE;
    END //
    DELIMITER ;

错误＃1054-“新建”中的未知列“ program_id”

2 个答案: