Question

我正在努力解决一个群体中的多数群体问题。举个例子，让我们说我的表看起来像这样：

+--------------------------------------------------+
|   city      |  car_colour |  car_type |  qty   |
+--------------------------------------------------+
| ------------------------------------------------ |
| manchester  |  Red        |  Sports   |  7       |
| manchester  |  Red        |  4x4      |  9       |
| manchester  |  Blue       |  4x4      |  8       |
| london      |  Red        |  Sports   |  2       |
| london      |  Blue       |  4x4      |  3       |
| leeds       |  Red        |  Sports   |  5       |
| leeds       |  Blue       |  Sports   |  6       |
| leeds       |  Blue       |  4X4      |  1       |
+--------------------------------------------------+

我试图找到一个纯粹的SQL解决方案，以便我可以看到：在每个城市，哪种颜色的汽车数量最多。

我能做到：

select city, cars, sum(qty)
from table
group by city, cars

得到：

+------------+------+----+
| manchester | red  | 16 |
| manchester | blue |  8 |
| london     | red  |  2 |
| london     | blue |  3 |
| leeds      | red  |  5 |
| leeds      | blue |  7 |
+------------+------+----+

但无论如何我可以使用子查询来获得结果的最大值，这将返回每个城市的最大颜色，因此结果将显示：

+------------+------+
| manchester | red  |
| london     | blue |
| leeds      | blue |
+------------+------+

我可以在我的Python脚本中进行这些计算，但更喜欢纯SQL解决方案。

希望这是有道理的，感谢您提前提供任何帮助：）

托米

Answer 1

select distinct p.city, p.car_colour,sq.qty as qty
from         ( select t.car_colour,t.city, sum(t.qty) as qty
                from table1 t
                group by t.car_colour,t.city
             )p
join         ( select q.city,max(q.qty) qty from
                  ( select t.car_colour,t.city, sum(t.qty) as qty
                   from table1 t
                   group by t.car_colour,t.city
                  )q
                 group by q.city
              )sq
on p.city=sq.city and p.qty=sq.qty

SQL FIDDLE DEMO

Answer 2

这样可行，但可能会根据您使用的特定数据库进行改进：

select t.city, t.car_colour, a.qty
from table1 t
join (
  select city, max(qty) qty 
  from (
    select city, car_colour, sum(qty) qty 
    from table1 
    group by city, car_colour
  ) x group by city
) a on t.city = a.city 
group by t.city, t.car_colour, a.qty
having sum(t.qty) = a.qty
order by t.city desc;

Sample SQL Fiddle

Answer 3

如果您使用MS SQL：

DECLARE @t TABLE
    (
      city NVARCHAR(MAX) ,
      color NVARCHAR(MAX) ,
      qty INT
    )

INSERT  INTO @t
VALUES  ( 'manchester', 'Red', 7 ),
        ( 'manchester', 'Red', 9 ),
        ( 'manchester', 'Blue', 8 ),
        ( 'london', 'Red', 2 ),
        ( 'london', 'Blue', 3 ),
        ( 'leeds', 'Red', 5 ),
        ( 'leeds', 'Blue', 6 ),
        ( 'leeds', 'Blue', 1 )


SELECT  city , color
FROM    ( SELECT    city ,
                    color ,
                    SUM(qty) AS q ,
                    ROW_NUMBER() OVER ( PARTITION BY city ORDER BY SUM(qty) DESC ) AS rn
          FROM      @t
          GROUP BY  city , color
        ) t
WHERE   rn = 1

输出：

city        color
leeds       Blue
london      Blue
manchester  Red

sql group by sub group

3 个答案: