0
A
回答
1
這有點凌亂,但它在這裏。所有的
首先,我使用的秩()函數來知道的位置(1,2或3)ErrorField的。所有這一切,我使用該數字爲樞軸。 你需要兩個樞軸和它們之間的連接。
WITH AuxTable (Data_Error_Key, ErrorField, ErrorValue, NumeroError)
AS
(
SELECT Data_Error_Key, ErrorField, ErrorValue, RANK() OVER (PARTITION BY Data_Error_Key ORDER BY Data_Error_Key, ErrorField)
FROM dbo.TempTable
)
SELECT TablaErrorField.Data_Error_Key, ErrorField1, ErrorValue1,ErrorField2,ErrorValue2, ErrorField3,
ErrorValue3
FROM
(
SELECT Data_Error_Key, [1] as ErrorField1, [2] as ErrorField2, [3] as ErrorField3
FROM (
SELECT Data_Error_Key,NumeroError, ErrorField
FROM AuxTable) P
PIVOT
(
MAX (ErrorField)
FOR NumeroError IN ([1], [2], [3])
) AS pvt) As TablaErrorField
INNER JOIN
(
SELECT Data_Error_Key, [1] as ErrorValue1, [2] as ErrorValue2, [3] as ErrorValue3
FROM (
SELECT Data_Error_Key,NumeroError, ErrorValue
FROM AuxTable) P
PIVOT
(
MAX (ErrorValue)
FOR NumeroError IN ([1], [2], [3])
) AS pvt) as TablaErrorValue
ON TablaErrorField.Data_Error_Key= TablaErrorValue.Data_Error_Key
這一切都假設您只需要3對ErrorField/ErrorValue。否則,你應該看看我的回答this的問題。
1
我不知道,如果你能做到這一點與SQL Server PIVOT功能。該函數總是假設並需要某種數值的集合函數(COUNT,AVG) - 您不能僅將行轉換爲列。
在你的情況,如果你事先知道你的錯誤的字段,你可以做這樣的事情:
SELECT
ErrorKey,
[Field1] AS Field1, [Field2] AS Field2, [Field10] AS Field10,
[Field11] AS Field11, [Field13] as Field13, [Field14] as Field14,
[Field15] as Field15, [Field21] as field21
FROM
(SELECT ErrorKey, ErrorField, ErrorValue
FROM Errors) e
PIVOT
(
COUNT (ErrorValue)
FOR ErrorField IN
([Field1], [Field2], [Field10], [Field11], [Field13],
[Field14], [Field15], [Field21])
) AS pvt
ORDER BY pvt.ErrorKey
這將產生一個輸出這樣的事情:
ErrorKey Field1 Field2 Field10 Field11 Field13 Field14 Field15 field21
1 1 1 1 0 0 0 0 0
2 0 0 0 1 1 1 0 0
3 0 0 0 0 0 0 1 1
這將使你爲每個錯誤鍵在特定字段中的錯誤計數。