如果与内部查询一起使用，SQL server pivot count函数将无法正常工作

我认为我的SQL server pivot函数中有一个bug，因为我找不到任何其他解释

我正在运行一个简单的pivot，并使用MSDN中显示的确切语法。 但是pivot答案显示了所有列的完全相同的数字，其值等于所有周的计数！！！ （左边是查询结果，右边是我想要的）

我的问题是：

SELECT
  *
FROM (SELECT
    r.cutomer_id
   ,c.[Week]
   ,r.id
  FROM r
  JOIN c
    ON r.Create_date = c.Date
  WHERE Is_ride = 1
  AND ((Create_date_int BETWEEN 20190302 AND 20190319)
  OR (Create_date_int BETWEEN 20190406 AND 20190426))) p
PIVOT
(
COUNT(id)
FOR [Week] IN
([9], [10], [11], [12], [14], [15], [16], [17])
) AS pvt

下面是一些表示“p”输出的测试数据（我刚刚更改了id号，周数与您从查询中得到的相同）

因此，我认为我的查询有问题，但后来我将查询更改为：

SELECT
  r.cutomer_id
 ,c.[Week]
 ,r.id INTO #t
FROM r
JOIN c
  ON r.Create_date = c.Date
WHERE Is_ride = 1
AND ((Create_date_int BETWEEN 20190302 AND 20190319)
OR (Create_date_int BETWEEN 20190406 AND 20190426))

SELECT
  *
FROM #t
PIVOT
(
COUNT(id)
FOR [Week] IN
([9], [10], [11], [12], [14], [15], [16], [17])
) AS pvt

而且效果很好！ 另外，如果我从select中删除r.id列，并将其更改为

count（week）

它工作正常

如果我只将我的

where

更改为

   WHERE Is_ride = 1
    AND ((Create_date_int BETWEEN 20190302 AND 20190319)
    OR (Create_date_int BETWEEN 20190406 AND 20190426))
    and passenger_id in (43551,12032,136019)
            ) p

它很好用

---

有人能给我一个解释吗？

我不知道你为什么会在图像上看到结果。可能丢失了一些信息。 下面是一个MVCE，其中的代码被注释以创建更大的测试数据集

CREATE TABLE r(
    id int identity,
    customer_id int, 
    create_date date, 
    Create_date_int AS CONVERT( int, CONVERT( char(8), create_date, 112)),
    is_ride bit)
INSERT INTO r(customer_id, create_date, is_ride)
SELECT customer_id, '20190307', 1
FROM (VALUES(1),(2),(3),(4),(5),(6),(7),(8),(9),(10))x(customer_id)
--SELECT TOP 100000
--       ABS(CHECKSUM(NEWID())) % 10,
--       DATEADD( dd, ABS(CHECKSUM(NEWID())) % 120, '2019'),
--       1
--FROM sys.all_columns a, sys.all_columns b;

CREATE TABLE c(
    [Date]  date,
    [Week]  AS DATEPART( wk, [Date])
)
INSERT INTO c([Date])
SELECT  DATEADD( dd, ROW_NUMBER() OVER( ORDER BY (SELECT NULL))-1, '2019')
FROM sys.all_columns;

从这里，我可以运行您最初的查询，得到预期的结果，在第10周只需1个

我还可以使用交叉表运行替代查询，这是我首选的数据透视方法

SELECT
    r.customer_id
    ,COUNT( CASE WHEN c.[Week] =  9 THEN r.id END) AS wk9
    ,COUNT( CASE WHEN c.[Week] = 10 THEN r.id END) AS wk10
    ,COUNT( CASE WHEN c.[Week] = 11 THEN r.id END) AS wk11
    ,COUNT( CASE WHEN c.[Week] = 12 THEN r.id END) AS wk12
    ,COUNT( CASE WHEN c.[Week] = 14 THEN r.id END) AS wk14
    ,COUNT( CASE WHEN c.[Week] = 15 THEN r.id END) AS wk15
    ,COUNT( CASE WHEN c.[Week] = 16 THEN r.id END) AS wk16
    ,COUNT( CASE WHEN c.[Week] = 17 THEN r.id END) AS wk17
FROM r
JOIN c ON r.Create_date = c.Date
WHERE Is_ride = 1
AND (Create_date_int BETWEEN 20190302 AND 20190319
OR Create_date_int BETWEEN 20190406 AND 20190426) 
GROUP BY r.customer_id
ORDER BY r.customer_id;

这也提供了正确的信息

这两种方法都可以通过预聚集来提高性能

WITH ctePreAggregate AS(
    SELECT
      r.customer_id
     ,c.[Week]
     ,COUNT(*) AS weeklycount
    FROM r
    JOIN c
      ON r.Create_date = c.Date
    WHERE Is_ride = 1
    AND (Create_date_int BETWEEN 20190302 AND 20190319
    OR Create_date_int BETWEEN 20190406 AND 20190426)
    GROUP BY r.customer_id
            ,c.[Week]
)
SELECT
  *
FROM ctePreAggregate
PIVOT
(
SUM(weeklycount)
FOR [Week] IN
([9], [10], [11], [12], [14], [15], [16], [17])
) AS pvt
ORDER BY customer_id;


WITH ctePreAggregate AS(
    SELECT
      r.customer_id
     ,c.[Week]
     ,COUNT(*) AS weeklycount
    FROM r
    JOIN c
      ON r.Create_date = c.Date
    WHERE Is_ride = 1
    AND (Create_date_int BETWEEN 20190302 AND 20190319
    OR Create_date_int BETWEEN 20190406 AND 20190426)
    GROUP BY r.customer_id
            ,c.[Week]
)
SELECT
    customer_id
    ,SUM( CASE WHEN [Week] =  9 THEN weeklycount ELSE 0 END) AS wk9
    ,SUM( CASE WHEN [Week] = 10 THEN weeklycount ELSE 0 END) AS wk10
    ,SUM( CASE WHEN [Week] = 11 THEN weeklycount ELSE 0 END) AS wk11
    ,SUM( CASE WHEN [Week] = 12 THEN weeklycount ELSE 0 END) AS wk12
    ,SUM( CASE WHEN [Week] = 14 THEN weeklycount ELSE 0 END) AS wk14
    ,SUM( CASE WHEN [Week] = 15 THEN weeklycount ELSE 0 END) AS wk15
    ,SUM( CASE WHEN [Week] = 16 THEN weeklycount ELSE 0 END) AS wk16
    ,SUM( CASE WHEN [Week] = 17 THEN weeklycount ELSE 0 END) AS wk17
FROM ctePreAggregate
GROUP BY customer_id
ORDER BY customer_id;

---

唯一的问题是，您需要在列列表中添加一系列ISNULL（），以便为PIVOT查询显示null而不是零。

“它工作正常！！！”我非常怀疑它工作正常，因为SQL表/结果集按SQL定义是无序的，所以使用

TOP

和

ORDER by

是毫无意义的。。事实上，它可能导致返回非确定性（随机）结果。。要始终获得确定性（固定）结果，您应该至少包含一个列，该列具有主键、唯一键，或者您知道该列具有（几乎）唯一的值，如datetime，按顺序为7分秒，假设当您开始大量删除或更新内容，并且表/索引碎片化成为一件事情时，它将不再正常工作。。我警告过你，如果你想继续用你的查询结果玩俄罗斯轮盘赌，那么这是你的选择。对于任何在这里帮助你的人，你需要提供一个再现问题或实际执行计划（XML）的MVCE。当然，如果将完全相同的行插入临时表而不是派生表，PIVOT不应返回不同的结果。可能是数据库中的某种损坏，一个bug，或者你的描述库中缺少什么，Martin。我们有非常敏感的数据和安全的服务器，很难创建MVCE。这些是我使用的完全相同的查询，我得到了完全不同的结果！如果没有任何解释，我想我的数据库可能有问题！我会检查/重建索引，或者尝试使用查询来缩小问题范围。

WITH ctePreAggregate AS(
    SELECT
      r.customer_id
     ,c.[Week]
     ,COUNT(*) AS weeklycount
    FROM r
    JOIN c
      ON r.Create_date = c.Date
    WHERE Is_ride = 1
    AND (Create_date_int BETWEEN 20190302 AND 20190319
    OR Create_date_int BETWEEN 20190406 AND 20190426)
    GROUP BY r.customer_id
            ,c.[Week]
)
SELECT
  *
FROM ctePreAggregate
PIVOT
(
SUM(weeklycount)
FOR [Week] IN
([9], [10], [11], [12], [14], [15], [16], [17])
) AS pvt
ORDER BY customer_id;


WITH ctePreAggregate AS(
    SELECT
      r.customer_id
     ,c.[Week]
     ,COUNT(*) AS weeklycount
    FROM r
    JOIN c
      ON r.Create_date = c.Date
    WHERE Is_ride = 1
    AND (Create_date_int BETWEEN 20190302 AND 20190319
    OR Create_date_int BETWEEN 20190406 AND 20190426)
    GROUP BY r.customer_id
            ,c.[Week]
)
SELECT
    customer_id
    ,SUM( CASE WHEN [Week] =  9 THEN weeklycount ELSE 0 END) AS wk9
    ,SUM( CASE WHEN [Week] = 10 THEN weeklycount ELSE 0 END) AS wk10
    ,SUM( CASE WHEN [Week] = 11 THEN weeklycount ELSE 0 END) AS wk11
    ,SUM( CASE WHEN [Week] = 12 THEN weeklycount ELSE 0 END) AS wk12
    ,SUM( CASE WHEN [Week] = 14 THEN weeklycount ELSE 0 END) AS wk14
    ,SUM( CASE WHEN [Week] = 15 THEN weeklycount ELSE 0 END) AS wk15
    ,SUM( CASE WHEN [Week] = 16 THEN weeklycount ELSE 0 END) AS wk16
    ,SUM( CASE WHEN [Week] = 17 THEN weeklycount ELSE 0 END) AS wk17
FROM ctePreAggregate
GROUP BY customer_id
ORDER BY customer_id;