基于SQL的选择独特的价值

Question

表：

Name | Value | Prize 
Race1 32  5 
Race1 22  5 
Race1 30  5 
Race2 11  5 
Race2 12  5 
Race2 31  5 

我想要的查询，当我定义Value 20+，独特的种族（Race1，RACE2）将被选中我的结果的基础上，其Value更高。应该总结奖金。

所以，当我做metaquery：select sum(Prize) biggestValueUniqueRaces where value > -20 我得到：

查询返回：

Prize 
    10 

凡考虑到这2行：

Name | Value | Prize 
Race1 32  5 
Race2 31  5 

Answer 1

编辑再评论

SELECT Sum(Races.Prize) As Total 
FROM Races INNER JOIN (SELECT r.[Name], Max(r.[Value]) AS MaxOfValue 
FROM Races r WHERE [Value]>20 
GROUP BY r.[Name]) AS q 
ON (Races.Value = q.MaxOfValue) AND (Races.Name = q.Name); 

或者

SELECT SUM(races.prize) AS SP 
FROM races 
WHERE races.id IN (
    SELECT id 
    FROM races r 
    WHERE r.name = races.name 
    AND VALUE = (
     SELECT Max(VALUE) 
     FROM races s 
     WHERE s.[name] = r.name 
     AND VALUE > 20)) 

Answer 2

你得先确定所有的比赛，对于其值等于具有相同名称的比赛的最大值。

SELECT 
    [r1].[Name], 
    [r1].[Value], 
    [r1].[Prize] 
    FROM [Races] [r1] 
    WHERE [r1].[Value] > 20 
    AND [r1].[Value] = 
     (SELECT MAX([r2].[Value]) 
     FROM [Races] [r2] 
     WHERE [r2].[Name] = [r1].[Name]) 

这可能会给多个种族具有相同的名称，价值和奖品。所以它们分组：

SELECT 
    [r1].[Name], 
    [r1].[Value], 
    [r1].[Prize] 
    FROM [Races] [r1] 
    WHERE [r1].[Value] > 20 
    AND [r1].[Value] = 
     (SELECT MAX([r2].[Value]) 
     FROM [Races] [r2] 
     WHERE [r2].[Name] = [r1].[Name]) 
    GROUP BY 
    [r1].[Name], 
    [r1].[Value], 
    [r1].[Prize] 

再总结奖品：

SELECT 
    SUM([sub].[Prize]) AS [TotalPrizes], 
    COUNT(*) AS [NumberOfRaces] 
FROM 
(
    SELECT 
    [r1].[Name], 
    [r1].[Value], 
    [r1].[Prize] 
    FROM [Races] [r1] 
    WHERE [r1].[Value] > 20 
    AND [r1].[Value] = 
     (SELECT MAX([r2].[Value]) 
     FROM [Races] [r2] 
     WHERE [r2].[Name] = [r1].[Name]) 
    GROUP BY 
    [r1].[Name], 
    [r1].[Value], 
    [r1].[Prize] 
) AS [sub] 

Here is a Fiddle。 （我将服务器类型设置为MS SQL 2008，但我认为这不会成为这种情况/查询的问题）。