Question

我有以下数据，我想根据轮回结果进行调整并获得计数。

DECLARE @tempMusicSchoolStudent TABLE
(school VARCHAR(50),
 studentname VARCHAR(50),
 instrumentname VARCHAR(255),
 expertise INT)

 INSERT INTO @tempMusicSchoolStudent(school, studentname, instrumentname, expertise)
 SELECT 'Foster','Matt','Guitar','10'
 UNION
 SELECT 'Foster','Jimmy','Guitar','5'
 UNION
 SELECT 'Foster','Jimmy','Keyboard','8'
 UNION
 SELECT 'Foster','Ryan','Keyboard','9' 
 UNION
 SELECT 'Midlothean','Kyle','Keyboard','10'
 UNION
 SELECT 'Midlothean','Mary','Guitar','4'
 UNION
 SELECT 'Midlothean','Mary','Keyboard','7'

原始数据：

我希望结果看起来像下面的数据....

我使用下面的sql查询获得了这些数据。这个查询的问题是我有一个动态数量的工具（为简单起见，我在这个例子中只显示了2个）。我想使用pivot，因为它将是更干净的动态sql。否则，我必须动态地将表连接到每个工具的自身。

SELECT 
    t.school, t.instrumentname, t.expertise,
    t1.instrumentname, t1.expertise,
    COUNT(DISTINCT t.studentname) [DistinctStudentCount]
FROM 
    @tempMusicSchoolStudent t
LEFT JOIN 
    @tempMusicSchoolStudent t1 ON t1.school = t.school 
                               AND t1.studentname = t.studentname 
                               AND t.instrumentname <> t1.instrumentname
GROUP BY 
    t.school, t.instrumentname, t.expertise, t1.instrumentname, t1.expertise
ORDER BY 
    t.school, t.instrumentname, t.expertise, t1.instrumentname, t1.expertise

如果有人对如何以更清洁的方式做到这一点有任何想法，而不是动态地将表连接到自身，那将非常感激。感谢。

Answer 1

您只需要条件聚合：

SELECT t.school, t.instrumentname, t.expertise, t.instrumentname, 
       COUNT(DISTINCT t.studentname) as DistinctStudentCount
FROM @tempMusicSchoolStudent t
GROUP BY t.school, t.instrumentname, t.expertise, t.instrumentname;

您的行包含NULL个值。目前尚不清楚这些来自哪里。你的问题集中在一些“转动”的概念，似乎你只需要聚合。但它没有解释NULL行的来源。

Answer 2

您可以尝试将其设置为多重乐器的动态。 Refer

;with cte
as
(
SELECT * from
(SELECT * FROM  @tempMusicSchoolStudent t) x
PIVOT
(MAX(expertise) FOR instrumentname in ([Guitar], [Keyboard])) y
)

SELECT school, studentname, 
 expertise = case when Guitar is not null then 'Guitar' else NULL end,
 Guitar AS instrumentname,
 expertise = case when Keyboard is not null then 'Keyboard' else NULL end,
 Keyboard AS instrumentname,
 count(distinct studentname) AS [DistinctStudentCount]
from cte
group by school,studentname, Guitar, Keyboard

输出：

Foster          Jimmy   Guitar  5     Keyboard  8      1
Foster          Matt    Guitar  10    NULL      NULL   1
Foster          Ryan    NULL    NULL  Keyboard  9      1
Midlothean      Kyle    NULL    NULL  Keyboard  10     1
Midlothean      Mary    Guitar  4     Keyboard  7      1

Answer 3

这是我正在寻找的解决方案，我不得不使用unpivot + pivot。

我正在努力解决的问题是为正在旋转的列选择多个值，而不是最大值。

因此，在这种情况下，我想在给定的“仪器专业知识”栏目下获得多个“专业知识”数字。不仅仅是该仪器的最大专业知识。

理解解决方案的第一个关键是pivot语句正在对所选列进行隐式组。因此，为了在您的透视列下实现多个值，您必须通过包含某种类型的dense_rank / rank / row_number来保持要分组的列的完整性。这基本上表示您正在旋转的列的值的变化，然后由枢轴正在执行的隐式组中包含，这导致在透视列中获取多个值，而不仅仅是最大值。

因此，在下面的代码中，“expertisenum”专栏保持专业知识数据的完整性。

DECLARE @tempMusicSchoolStudent TABLE
(school VARCHAR(50),
 studentname VARCHAR(50),
 instrumentname VARCHAR(255),
 expertise INT)

INSERT INTO @tempMusicSchoolStudent(school, studentname, instrumentname, expertise)
SELECT 'Foster','Matt','Guitar','10'
UNION
SELECT 'Foster','Jimmy','Guitar','5'
UNION
SELECT 'Foster','Jimmy','Keyboard','8'
UNION
SELECT 'Foster','Ryan','Keyboard','9' 
UNION
SELECT 'Midlothean','Kyle','Keyboard','10'
UNION
SELECT 'Midlothean','Mary','Guitar','4'
UNION
SELECT 'Midlothean','Mary','Keyboard','7'



SELECT school, [Guitar expertise], [Keyboard expertise], COUNT(*) [Count]
FROM
(
    SELECT school,[expertiseNum],
    CASE WHEN [Columns]='expertise' THEN instrumentname + ' expertise'
         END [Columns1], [Values] AS [Values1]
    FROM
    (
        SELECT school, studentname, instrumentname, DENSE_RANK() OVER(PARTITION BY school,instrumentname ORDER BY expertise) AS [expertiseNum],
        CONVERT(VARCHAR(255),expertise) AS [expertise]
        FROM @tempMusicSchoolStudent
    ) x
    UNPIVOT (
        [Values] FOR [Columns] IN ([expertise])
    ) unpvt
) p
PIVOT (
    MAX([Values1]) FOR [Columns1] IN ([Guitar expertise], [Keyboard expertise])
) pvt
GROUP BY school,[Guitar expertise], [Keyboard expertise]

T-SQL数据透视表，其中包含对透视结果的计数

3 个答案: