如何从SQL中的字段计算相同的评分

Question

我在计算SQL中的评分时遇到问题。这是我的数据：

data

 CREATE TABLE `restaurant` (
  `id_restaurant` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(50) DEFAULT NULL,
  PRIMARY KEY (`id_restaurant`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=latin1;

insert  into `restaurant`(`id_restaurant`,`name`) values (1,'Mc Donald');
insert  into `restaurant`(`id_restaurant`,`name`) values (2,'KFC');

    CREATE TABLE `user` (
  `id_user` int(11) NOT NULL AUTO_INCREMENT,
  `userName` varchar(50) DEFAULT NULL,
  PRIMARY KEY (`id_user`)
) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=latin1;

insert  into `user`(`id_user`,`userName`) values (1,'Audey');


    CREATE TABLE `factors` (
  `factor_id` int(11) NOT NULL AUTO_INCREMENT,
  `factor_clean` int(11) NOT NULL DEFAULT '0',
  `factor_delicious` int(11) NOT NULL DEFAULT '0',
  `id_restaurant` int(11) DEFAULT NULL,
  `id_user` int(11) DEFAULT NULL,
  PRIMARY KEY (`factor_id`),
  KEY `id_restaurant` (`id_restaurant`),
  KEY `id_user` (`id_user`),
  CONSTRAINT `factors_ibfk_1` FOREIGN KEY (`id_restaurant`) REFERENCES `restaurant` (`id_restaurant`),
  CONSTRAINT `factors_ibfk_2` FOREIGN KEY (`id_user`) REFERENCES `user` (`id_user`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=latin1;

    insert  into `factors`(`factor_id`,`factor_clean`,`factor_delicious`,`id_restaurant`,`id_user`) values (1,1,5,1,1);
insert  into `factors`(`factor_id`,`factor_clean`,`factor_delicious`,`id_restaurant`,`id_user`) values (2,0,5,1,1);
insert  into `factors`(`factor_id`,`factor_clean`,`factor_delicious`,`id_restaurant`,`id_user`) values (3,1,5,1,1);
insert  into `factors`(`factor_id`,`factor_clean`,`factor_delicious`,`id_restaurant`,`id_user`) values (4,3,3,1,1);

结果应该是这样，显示rating_clean，rating_delicious和rating_clean字段中的所有评分（1,2,3,4,5）及其计数/ p>

感谢您的帮助。

但是我得到的结果

SELECT COUNT(`factor_clean`+`factor_delicious`),'1' AS rating_1 FROM `factors` WHERE 1 GROUP BY `id_restaurant`

result not should like this

结果不应该那样， 我的问题是，如何只选择factor_clean和factor_delicious，其中factor_clean = 1和factor_delicious = 1

Answer 1

使用union all取消数据透视，然后进行汇总：

select id_restaurant, rating, count(*)
from ((select r.id_restaurant, r.rating_clean as rating, r.date
       from ratings r
      ) union all
      (select r.id_restaurant, r.rating_delicious, r.date
       from ratings r
      ) union all
      (select r.id_restaurant, r.rating_clean2, r.date
       from ratings r
      ) 
     ) r
group by id_restaurant, rating
order by id_restaurant, rating;

Answer 2

例如，这是针对列为rating_delicious和rating_clean（只有一个！）的表的解决方案：

首先，您应该创建其他表，我称它为因素：

CREATE TABLE `factors` (
 `factor_id` int(11) NOT NULL AUTO_INCREMENT,
 `factor_clean` int(11) NOT NULL DEFAULT '0',
 `factor_delicious` int(11) NOT NULL DEFAULT '0',
 PRIMARY KEY (`factor_id`)
)

接下来添加两条记录：

INSERT INTO `factors` (`factor_id`, `factor_clean`, `factor_delicious`) VALUES (NULL, '1', '0'), (NULL, '0', '1');

现在您可以加入此表并获取结果：

SELECT x.id_restaurant
     , (x.rating_clean * f.factor_clean) + (x.rating_delicious * f.factor_delicious) AS rating
     , count(*) 
  FROM your_table x
  JOIN factors f
 WHERE 1 
 GROUP 
    BY x.id_restaurant
     , rating

为了使用下一个列（rating_third），您应该将factor_third列到factors，并在此列中插入带有1的新行，最后添加类似your_table.rating_third*factors.factor_third合计为SELECT