1、题目描述
Write an SQL query that reports the best seller by total sales price, If there is a tie, report them all.
The query result format is in the following example:
Product table:
| product_id | product_name | unit_price |
|---|---|---|
| 1 | S8 | 1000 |
| 2 | G4 | 800 |
| 3 | iPhone | 1400 |
Sales table:
| seller_id | product_id | buyer_id | sale_date | quantity | price |
|---|---|---|---|---|---|
| 1 | 1 | 1 | 2019-01-21 | 2 | 2000 |
| 1 | 2 | 2 | 2019-02-17 | 1 | 800 |
| 2 | 2 | 3 | 2019-06-02 | 1 | 800 |
| 3 | 3 | 4 | 2019-05-13 | 2 | 2800 |
Result table:
| seller_id |
|---|
| 1 |
| 3 |
Both sellers with id 1 and 3 sold products with the most total price of 2800.
来源：力扣（LeetCode）
2、解题思路
其实和上一题都是差不多的，这次换一种解法
1# 首先，对每个销售员进行出售价格求和，然后倒序
select seller_id ,sum(price) as p
from Sales
group by seller_id
order by price desc
2# 然后，增加一列排名列 参考 分数排名
select seller_id ,p,@j:=@j+(@i<>(@i:=p)) as rank
3# 最后，就是找出rank=1的记录
3、提交记录
select seller_id
from(
select seller_id ,p,@j:=@j+(@i<>(@i:=p)) as rank
from(
select seller_id ,sum(price) as p
from Sales
group by seller_id
order by price desc)b,(select @i:=0,@j:=0)a)c
where rank=1
order by seller_id
