Трехколонный SQL PIVOT

Question

Как создать сводную таблицу данных, которая выглядит следующим образом, ИСПОЛЬЗУЯ команду SQL PIVOT?

id           |    field     |   value
---------------------------------------
1            |    year      |   2011
1            |    month     |   August
2            |    year      |   2009
1            |    day       |   21
2            |    day       |   31
2            |    month     |   July
3            |    year      |   2010
3            |    month     |   January
3            |    day       |   NULL

На что-то похожее на это:

id  |  year  |  month  |  day
-----------------------------
1     2011     August    21
2     2010      July     31
3     2009     January   NULL

Answer 1

Попробуйте что-то вроде этого:

DECLARE @myTable AS TABLE([ID] INT, [Field] VARCHAR(20), [Value] VARCHAR(20))
INSERT INTO @myTable VALUES ('1', 'year', '2011')
INSERT INTO @myTable VALUES ('1', 'month', 'August')
INSERT INTO @myTable VALUES ('2', 'year', '2009')
INSERT INTO @myTable VALUES ('1', 'day', '21')
INSERT INTO @myTable VALUES ('2', 'day', '31')
INSERT INTO @myTable VALUES ('2', 'month', 'July')
INSERT INTO @myTable VALUES ('3', 'year', '2010')
INSERT INTO @myTable VALUES ('3', 'month', 'January')
INSERT INTO @myTable VALUES ('3', 'day', NULL)

SELECT [ID], [year], [month], [day]
FROM
(
      SELECT [ID], [Field], [Value] FROM @myTable
) t
PIVOT
(
    MIN([Value]) FOR [Field] IN ([year], [month], [day]) 
) AS pvt
ORDER BY pvt.[year] DESC

Что даст результаты:

ID  year    month   day
1   2011    August  21
3   2010    January NULL
2   2009    July    31

Answer 2

;WITH DATA(id,field,value) AS
(
SELECT 1,'year','2011' UNION ALL
SELECT 1,'month','August' UNION ALL
SELECT 2,'year','2009' UNION ALL
SELECT 1,'day ','21' UNION ALL
SELECT 2,'day ','31' UNION ALL
SELECT 2,'month','July' UNION ALL
SELECT 3,'year','2010' UNION ALL
SELECT 3,'month','January' UNION ALL
SELECT 3,'day ',NULL
)
SELECT id,
       year,
       month,
       day
FROM   DATA PIVOT (MAX(value) FOR field IN ([year], [month], [day])) AS Pvt  

Answer 3

SELECT
id,
MAX(CASE WHEN RK=3 THEN VAL ELSE '' END) AS "YEAR",
MAX(CASE WHEN RK=2 THEN VAL ELSE '' END) AS "MONTH",
MAX(CASE WHEN RK=1 THEN VAL ELSE '' END) AS "DAY"

FROM 
(
SELect
ID,
ROW_NUMBER() OVER(PARTITION BY ID ORDER BY YEAR1 ASC) RK,
VAL
FROM TEST3)A
GROUP BY 1
ORDER BY 1;