PIVOT 没有问题，但对我来说，最简单和最高效的方法是执行Cross Tab。语法不那么冗长，更便携，更容易理解。
首先是一些 DDL 和易于使用的示例数据。
SET NOCOUNT ON;
SET DATEFORMAT dmy; -- I need this because I'm American
-- DDL and easily consumable sample data
DECLARE @Results TABLE
(
[Date] DATE,
EventId INT,
[Name] VARCHAR(40), -- if indexed, go as narrow as possible
[Time] TIME,
Points INT,
INDEX uq_poc_results CLUSTERED([Name],[EventId]) -- a covering index is vital for a query like this
); -- note: ^^^ this bad clustered index candidate, I went this route for simplicity
INSERT @Results VALUES
('4/04/2019', 10538, 'Fred Flintstone', '22:27',10),
('24/04/2019',10538, 'Barney Rubble', '22:50',9),
('24/04/2019',10538, 'Micky Mouse ', '23:17',8),
('24/04/2019',10538, 'Yogi Bear', '23:54',7),
('24/04/2019',10538, 'Donald Duck', '2307',6),
('01/05/2019',10541, 'Barney Rubble', '21:58',10),
('01/05/2019',10541, 'Fred Flintstone', '22:00',9),
('01/05/2019',10541, 'Donald Duck', '23:39',8),
('01/05/2019',10541, 'Yogi Bear', '23:43',7),
('12/06/2019',10569, 'Fred Flintstone', '22:06',10),
('12/06/2019',10569, 'Barney Rubble', '22:22',9),
('12/06/2019',10569, 'Micky Mouse', '23:05',8),
('12/06/2019',10569, 'Donald Duck', '23:55',7);
请注意，我在 (Name,EventId) 上创建了一个聚集索引 - 我将使用一个涵盖您在现实世界中需要的列的非聚集索引。如果您有很多行，那么您将需要该索引。
基本交叉表
SELECT [Name] = r.[Name],
[24/04/2019] = MAX(CASE r.[Date] WHEN '24/04/2019' THEN r.Points ELSE 0 END),
[01/05/2019] = MAX(CASE r.[Date] WHEN '01/05/2019' THEN r.Points ELSE 0 END),
[12/06/2019] = MAX(CASE r.[Date] WHEN '12/06/2019' THEN r.Points ELSE 0 END)
FROM @Results AS r
GROUP BY r.[Name];
结果：
Name 24/04/2019 01/05/2019 12/06/2019
-------------------- ------------ ------------ ------------
Barney Rubble 9 10 9
Donald Duck 6 8 7
Fred Flintstone 0 9 10
Micky Mouse 8 0 8
Yogi Bear 7 7 0
要获得总数，我们可以将其包装在子查询中的逻辑中并添加如下所示的列：
SELECT
[Name] = piv.N,
[24/04/2019] = piv.D1,
[01/05/2019] = piv.D2,
[12/06/2019] = piv.D3,
Total = piv.D1+piv.D2+piv.D3
FROM
(
SELECT r.[Name],
MAX(CASE r.[Date] WHEN '24/04/2019' THEN r.Points ELSE 0 END),
MAX(CASE r.[Date] WHEN '01/05/2019' THEN r.Points ELSE 0 END),
MAX(CASE r.[Date] WHEN '12/06/2019' THEN r.Points ELSE 0 END)
FROM @Results AS r
GROUP BY r.[Name]
) AS piv(N,D1,D2,D3);
返回：
Name 24/04/2019 01/05/2019 12/06/2019 Total
------------------- ----------- ----------- ----------- -------
Barney Rubble 9 10 9 28
Donald Duck 6 8 7 21
Fred Flintstone 0 9 10 19
Micky Mouse 8 0 8 16
Yogi Bear 7 7 0 14
这不仅可以用很少的 SQL 为您提供所需的东西，而且您还可以从子查询中的预聚合中受益。与PIVOT 相比，这种方法的一个巨大好处是您可以如何在一个查询中进行多个聚合。以下是如何将这种方法用于多个聚合的两个示例；这首先使用标准 GROUP BY 两次，另一个使用窗口聚合函数(.. OVER (partition by, order by..)：
--==== Traditional Approach
SELECT
[Name] = piv.N,
[24/04/2019] = MAX(piv.D1),
[01/05/2019] = MAX(piv.D2),
[12/06/2019] = MAX(piv.D3),
Total = MAX(f.Ttl),
Avg1 = AVG(piv.D1), -- 1st date (24/04/2019)
Avg2 = AVG(piv.D2), -- 2nd date...
Avg3 = AVG(piv.D3), -- 3rd date...
TotalAvg = AVG(f.Ttl) ,
Mn = MIN(f.Ttl) ,
Mx = MAX(f.Ttl)
FROM
(
SELECT r.[Name],
MAX(CASE r.[Date] WHEN '24/04/2019' THEN r.Points ELSE 0 END),
MAX(CASE r.[Date] WHEN '01/05/2019' THEN r.Points ELSE 0 END),
MAX(CASE r.[Date] WHEN '12/06/2019' THEN r.Points ELSE 0 END)
FROM @Results AS r
GROUP BY r.[Name]
) AS piv(N,D1,D2,D3)
CROSS APPLY (VALUES(piv.D1+piv.D2+piv.D3)) AS f(Ttl)
GROUP BY piv.N;
--==== Leveraging Window Aggregates
SELECT
[Name] = piv.N,
[24/04/2019] = piv.D1,
[01/05/2019] = piv.D2,
[12/06/2019] = piv.D3,
Total = f.Ttl,
Avg1 = AVG(piv.D1) OVER(PARTITION BY piv.N ORDER BY (SELECT NULL)), -- 1st date (24/04/2019)
Avg2 = AVG(piv.D2) OVER(PARTITION BY piv.N ORDER BY (SELECT NULL)), -- 2nd date...
Avg3 = AVG(piv.D3) OVER(PARTITION BY piv.N ORDER BY (SELECT NULL)), -- 3rd date...
TotalAvg = AVG(f.Ttl) OVER(PARTITION BY piv.N ORDER BY (SELECT NULL)),
Mn = MIN(f.Ttl) OVER(PARTITION BY piv.N ORDER BY (SELECT NULL)),
Mx = MAX(f.Ttl) OVER(PARTITION BY piv.N ORDER BY (SELECT NULL))
FROM
(
SELECT r.[Name],
MAX(CASE r.[Date] WHEN '24/04/2019' THEN r.Points ELSE 0 END),
MAX(CASE r.[Date] WHEN '01/05/2019' THEN r.Points ELSE 0 END),
MAX(CASE r.[Date] WHEN '12/06/2019' THEN r.Points ELSE 0 END)
FROM @Results AS r
GROUP BY r.[Name]
) AS piv(N,D1,D2,D3)
CROSS APPLY (VALUES(piv.D1+piv.D2+piv.D3)) AS f(Ttl);
两者都返回：
Name 24/04/2019 01/05/2019 12/06/2019 Total Avg1 Avg2 Avg3 TotalAvg Mn Mx
----------------- ----------- ----------- ----------- ------ ------ ------ ------ ---------- ------ ------
Barney Rubble 9 10 9 28 9 10 9 28 28 28
Donald Duck 6 8 7 21 6 8 7 21 21 21
Fred Flintstone 0 9 10 19 0 9 10 19 19 19
Micky Mouse 8 0 8 16 8 0 8 16 16 16
Yogi Bear 7 7 0 14 7 7 0 14 14 14
要动态处理列，您需要查看：
Cross Tabs and Pivots, Part 2 - Dynamic Cross Tabs Jeff Moden。