【SQL练习题】排序

原表如下：

建表脚本：

CREATE TABLE districtproducts (   district varchar(255),   name varchar(255),   price int(255) );  INSERT INTO districtproducts VALUES ('东北', '橘子', 100); INSERT INTO districtproducts VALUES ('东北', '苹果', 50); INSERT INTO districtproducts VALUES ('东北', '葡萄', 50); INSERT INTO districtproducts VALUES ('东北', '柠檬', 30); INSERT INTO districtproducts VALUES ('关东', '柠檬', 100); INSERT INTO districtproducts VALUES ('关东', '菠萝', 100); INSERT INTO districtproducts VALUES ('关东', '苹果', 100); INSERT INTO districtproducts VALUES ('关东', '葡萄', 70); INSERT INTO districtproducts VALUES ('关西', '柠檬', 70); INSERT INTO districtproducts VALUES ('关西', '西瓜', 30); INSERT INTO districtproducts VALUES ('关西', '苹果', 20);

效果如下：

每个district中的水果按照价格降序排序，rank_1跳过排名，rank_2不跳过排名

三种方式实现：

select p.district,p.name,p.price, rank() over (partition by p.district order by p.price desc) as rank_1, dense_rank() over (partition by p.district order by p.price desc) as rank_2 from districtproducts p;

select  p1.district,p1.name,p1.price, (select count(p2.price) from districtproducts p2 where p1.district = p2.district and p2.price > p1.price) + 1 as rank_1, (select count(distinct p2.price) from districtproducts p2 where p1.district = p2.district and p2.price > p1.price) + 1 as rank_2 from districtproducts p1;

select  p1.district,p1.name,max(p1.price) as price,count(p2.price)+1 as rank_1,count(distinct p2.price)+1 as rank_2 from districtproducts p1 left outer join districtproducts p2 on p1.district = p2.district and p1.price < p2.price  group by p1.district,p1.name order by p1.district,price desc;

文章来源: 【SQL练习题】排序