Select 第一列不同的行和第二列的最大值

Question

我有以下 table:

Item           Prod        Company
1.00961.501    PR011798    ditto
1.00961.501    PR012042    ditto
1.00961.501    PR013442    Pika
1.00961.502    PR012043    ditto
1.00961.503    PR011959    ditto
1.00961.503    PR011669    Bulb
1.00961.507    PR014783    ditto
1.00961.507    PR012050    ditto

我想 select 所有 table 按 Item 分组，只取最大值 Prod。像这样：

Item           Prod        Company
1.00961.501    PR012042    ditto
1.00961.502    PR012043    ditto
1.00961.503    PR011959    ditto
1.00961.507    PR014783    ditto

我尝试了以下方法：

SELECT  DISTINCT Item, MAX(Prod)
FROM    DataBase
WHERE   Company = 'ditto'

但它给了我

Column 'Item' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.

如果我删除 MAX 子句，它 returns 没有错误，但是 Item 对每个 Prod 重复。

有什么想法吗？

编辑

我忘了在问题中添加 Where 子句。当我这样做并尝试使用 Group By 而不是 Distinct 时，我收到以下错误：

Column 'Company' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.

Answer 1

您的查询应该是这样的；

SELECT  Item, MAX(Prod),Company
FROM    DataBase
WHERE   Company = 'ditto'
GROUP BY Item,Company;

Answer 2

您需要对产品进行分区，然后 select 从该分区中提取一个。

您应该尝试下面的查询。

  select Item, Prod, Company from 
  (
    select *,
    (ROW_NUMBER() over (partition by Item order by Prod DESC)) as Row from [table_name] 
    where Company = 'ditto'
  ) t where Row = 1

Select 第一列不同的行和第二列的最大值

Select distinct rows by first column and max of second

sql

tsql

group-by

distinct

编辑