爲PIVOT源排序數據

Question

想象一下帶有屬性的表格（鍵值）和帶有批次的父表格。

LotId  SomeText 
----------- -------- 
1   Hello 
2   World 

AttributeId LotId  Val  Kind 
----------- ----------- -------- -------- 
1   1   Foo1  Kind1 
2   1   Foo2  Kind2 
3   2   Bar1  Kind1 
4   2   Bar2  Kind2 
5   2   Bar3  Kind3 

我使用UNPIVOT - PIVOT操作以獲取在表單中的數據：

LotId  SomeText AttributeId LotId  Kind1Val Kind  AttributeId LotId  Kind2Val Kind  AttributeId LotId  Kind3Val Kind 
----------- -------- ----------- ----------- -------- -------- ----------- ----------- -------- -------- ----------- ----------- -------- -------- 
1   Hello 1   1   Foo1  Kind1 2   1   Foo2  Kind2 NULL  NULL  NULL  NULL 
2   World 3   2   Bar1  Kind1 4   2   Bar2  Kind2 5   2   Bar3  Kind3 

如何選擇數據獨立性從屬性表中的值了。

不正確的結果的一個例子：

LotId  SomeText attributeid_1 LotId_1 Value_1 Kind_1 attributeid_2 LotId_2 Value_2 Kind_2 attributeid_3 LotId_3 Value_3 Kind_3 
----------- -------- ------------- -------- -------- -------- ------------- -------- -------- -------- ------------- -------- -------- -------- 
1   Hello 4    1  Foo1  Kind1 NULL   NULL  NULL  NULL  NULL   NULL  NULL  NULL 
2   World 1    2  Bar2  Kind2 3    2  Bar3  Kind3 2    2  Bar1  Kind8 

爲什麼？

由於Kind2Kind_1列中的文本和Kind3中的Kind_2。

SQL Fiddle

 
DECLARE @Lot TABLE (
LotId INT PRIMARY KEY IDENTITY, 
SomeText VARCHAR(8)) 

INSERT INTO @Lot 
VALUES ('Hello'), ('World') 

DECLARE @Attribute TABLE(
AttributeId INT PRIMARY KEY IDENTITY, 
LotId INT, 
Val VARCHAR(8), 
Kind VARCHAR(8)) 

INSERT INTO @Attribute VALUES 
(2, 'Bar2', 'Kind2'), 
(2, 'Bar1', 'Kind8'), 
(2, 'Bar3', 'Kind3'), 
(1, 'Foo1', 'Kind1') 


select * 
from 
(
    select LotId, 
     SomeText, 
     col+'_'+CAST(rn as varchar(10)) col, 
     value 
    from 
    (
     select l.LotId, 
      l.SomeText, 
      cast(a.AttributeId as varchar(8)) attributeid, 
      cast(a.LotId as varchar(8)) a_LotId, 
      a.Val, 
      a.Kind, 
      ROW_NUMBER() over(partition by l.lotid order by a.kind) rn 
     from @Lot l 
     left join @Attribute a 
      on l.LotId = a.LotId 
    ) src 
    cross apply 
    (
     values ('attributeid', attributeid),('LotId', a_LotId), ('Value', Val), ('Kind', Kind) 
    ) c (col, value) 
) d 
pivot 
(
    max(value) 
    for col in (attributeid_1, LotId_1, Value_1, Kind_1, 
       attributeid_2, LotId_2, Value_2, Kind_2, 
       attributeid_3, LotId_3, Value_3, Kind_3) 
) piv 

正確的結果的一個例子：

LotId  SomeText attributeid_Kind1 LotId_Kind1 Value_Kind1 Kind_Kind1 attributeid_Kind2 LotId_Kind2 Value_Kind2 Kind_Kind2 attributeid_Kind3 LotId_Kind3 Value_Kind3 Kind_Kind3 
----------- -------- ----------------- ----------- ----------- ---------- ----------------- ----------- ----------- ---------- ----------------- ----------- ----------- ---------- 
1   WithAll 1     1   Foo1  Kind1  2     1   Foo2  Kind2  3     1   Foo3  Kind3 
2   Hello NULL    NULL  NULL  NULL  10    2   Bar2  Kind2  NULL    NULL  NULL  NULL 
3   World NULL    NULL  NULL  NULL  NULL    NULL  NULL  NULL  12    3   Bar3  Kind3 

由於KindX是在Kind_x列等

SQL Fiddle

 
DECLARE @Lot TABLE (
LotId INT PRIMARY KEY IDENTITY, 
SomeText VARCHAR(8)) 

INSERT INTO @Lot 
VALUES ('WithAll'), ('Hello'), ('World') 

DECLARE @Attribute TABLE(
AttributeId INT PRIMARY KEY IDENTITY, 
LotId INT, 
Val VARCHAR(8), 
Kind VARCHAR(8)) 

INSERT INTO @Attribute VALUES 
(1, 'Foo1', 'Kind1'), 
(1, 'Foo2', 'Kind2'), 
(1, 'Foo3', 'Kind3'), 
(1, 'Foo4', 'Kind4'), 
(1, 'Foo5', 'Kind5'), 
(1, 'Foo6', 'Kind6'), 
(1, 'Foo7', 'Kind7'), 
(1, 'Foo8', 'Kind8'), 
(1, 'Foo9', 'Kind9'), 
(2, 'Bar2', 'Kind2'), 
(2, 'Bar1', 'Kind8'), 
(3, 'Bar3', 'Kind3') 

DECLARE @AttributesMask TABLE(
    Kind VARCHAR(8) 
) 

INSERT INTO @AttributesMask 
VALUES('Kind1'), ('Kind2'), ('Kind3'), ('Kind4'), ('Kind5'), ('Kind6'), ('Kind7'), ('Kind8') 

select * from(
    select LotId, 
     SomeText, 
     --col+'_'+CAST(rn as varchar(10)) col, 
     col+'_'+[Kind] col, 
     value 
    from 
    (
     select l.LotId, 
      l.SomeText, 
      cast(a.AttributeId as varchar(8)) attributeid, 
      cast(a.LotId as varchar(8)) a_LotId, 
      a.Val, 
      a.Kind 
      --, ROW_NUMBER() over(partition by l.[LotId] order by am.[Kind]) rn 
     FROM @AttributesMask AS am 
     LEFT join @Attribute a on [am].[Kind] = [a].[Kind] 
     LEFT JOIN @Lot l ON [a].[LotId] = [l].[LotId] 

    ) src 
    cross apply 
    (
     values ('attributeid', attributeid),('LotId', a_LotId), ('Value', Val), ('Kind', Kind) 
    ) c (col, value) 
) d PIVOT (max(value) for col in (
    attributeid_Kind1, LotId_Kind1, Value_Kind1, Kind_Kind1, 
    attributeid_Kind2, LotId_Kind2, Value_Kind2, Kind_Kind2, 
    attributeid_Kind3, LotId_Kind3, Value_Kind3, Kind_Kind3)) piv 
ORDER BY LotId 

爲了獲得正確的結果，我使用了掩碼來預先排列作爲PIVOT源的數據。如何做到這一點沒有面具？

參考的問題：How to replace a functional (many) OUTER APPLY (SELECT * FROM)

Answer 1

除非我的思念在你的解釋的東西，那麼你不需要AttributeMask。如果最終列名稱只是將是原來的列名，然後Kind值，那麼你可以使用：

select * 
from 
(
    select LotId, 
     SomeText, 
     col+'_'+Kind col, 
     value 
    from 
    (
     select l.LotId, 
      l.SomeText, 
      cast(a.AttributeId as varchar(8)) attributeid, 
      cast(a.LotId as varchar(8)) a_LotId, 
      a.Val, 
      a.Kind 
     from @Lot l 
     left join @Attribute a 
      on l.LotId = a.LotId 
    ) src 
    cross apply 
    (
     values ('attributeid', attributeid),('LotId', a_LotId), ('Value', Val), ('Kind', Kind) 
    ) c (col, value) 
) d 
pivot 
(
    max(value) 
    for col in (attributeid_Kind1, LotId_Kind1, Value_Kind1, Kind_Kind1, 
       attributeid_Kind2, LotId_Kind2, Value_Kind2, Kind_Kind2, 
       attributeid_Kind3, LotId_Kind3, Value_Kind3, Kind_Kind3) 
) piv; 

見SQL Fiddle with Demo。這給出了結果：

| LOTID | SOMETEXT | ATTRIBUTEID_KIND1 | LOTID_KIND1 | VALUE_KIND1 | KIND_KIND1 | ATTRIBUTEID_KIND2 | LOTID_KIND2 | VALUE_KIND2 | KIND_KIND2 | ATTRIBUTEID_KIND3 | LOTID_KIND3 | VALUE_KIND3 | KIND_KIND3 | 
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- 
|  1 | WithAll |     1 |   1 |  Foo1 |  Kind1 |     2 |   1 |  Foo2 |  Kind2 |     3 |   1 |  Foo3 |  Kind3 | 
|  2 | Hello |   (null) |  (null) |  (null) |  (null) |    10 |   2 |  Bar2 |  Kind2 |   (null) |  (null) |  (null) |  (null) | 
|  3 | World |   (null) |  (null) |  (null) |  (null) |   (null) |  (null) |  (null) |  (null) |    12 |   3 |  Bar3 |  Kind3 |