Question

我有一个名为 tbl_points 的表，其中包含以下列：

[key] identity
[fkey] int -> forign key from other table
[points] int -> number
[inputdate] datetime -> getdate()

和价值观如下：

key,fkey,points,inputdate

1,23,5,20170731
2,23,2,20170801
3,23,4,20170801
4,25,2,20170801
5,25,2,20170802
6,23,5,20170803
7,25,3,20170803
8,23,5,20170804

我正在执行这样的查询：

select fkey,sum(points) points,month(inputdate) mnd,year(inputdate) yy
from tbl_points 
group by fkey,month(inputdate) mnd,year(inputdate)
order by year(inputdate),month(inputdate) mnd,points

结果如下：

fkey,points,mnd,yy

23,14,8,2017
25,7,8,2017
25,5,7,2017

到目前为止一切顺利。现在我只想要每个月的前1 ，所以

23,14,8,2017
25,5,7,2017

我可以在代码中执行此操作，也可以使用临时表或游标在存储过程中执行此操作。

但也许有更简单的解决方案。有任何想法吗？还是更好的方法？

Answer 1

DECLARE @tbl_points TABLE
    (
      [key] INT ,
      [fkey] INT ,
      [points] INT ,
      [inputdate] DATETIME
    );

INSERT  INTO @tbl_points
VALUES  ( 1, 23, 5, '2017-07-31' ),
        ( 2, 23, 2, '2017-08-01' ),
        ( 3, 23, 4, '2017-08-01' ),
        ( 4, 25, 2, '2017-08-01' ),
        ( 5, 25, 2, '2017-08-02' ),
        ( 6, 23, 5, '2017-08-03' ),
        ( 7, 25, 3, '2017-08-03' ),
        ( 8, 23, 5, '2017-08-04' ); 


/* Your query */
SELECT  fkey ,
        SUM(points) points ,
        YEAR(inputdate) [year] ,
        MONTH(inputdate) [month]
FROM    @tbl_points
GROUP BY fkey ,
        MONTH(inputdate) ,
        YEAR(inputdate)
ORDER BY YEAR(inputdate) ,
        MONTH(inputdate) ,
        points;


/* Query you want */
SELECT  s.fkey ,
        s.points ,
        s.[year] ,
        s.[month]
FROM    ( SELECT    fkey ,
                    SUM(points) points ,
                    YEAR(inputdate) [year] ,
                    MONTH(inputdate) [month] ,
                    ROW_NUMBER() OVER ( PARTITION BY MONTH(inputdate) ORDER BY YEAR(inputdate) , MONTH(inputdate) , SUM(points) ASC ) [Row]
          FROM      @tbl_points
          GROUP BY  fkey ,
                    MONTH(inputdate) ,
                    YEAR(inputdate)
        ) AS s
WHERE   s.Row = 1;

结果：

使用group by获得独特的月份

1 个答案: