Question

想象一下带有属性的表（键值）和带有Lots的父表。

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

我正在使用UNPIVOT - PIVOT操作来获取以下形式的数据：

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

如何选择远离属性表值的数据独立性。

错误结果的示例：

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

为什么？

由于Kind2列中的Kind_1文字和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

正确结果的示例：

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

由于KindX位于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

为了得到正确的结果，我使用了掩码来预先安排作为PIVOT源的数据。如果没有面具怎么办？

参考问题：How to replace a functional (many) OUTER APPLY (SELECT * FROM)

Answer 1

除非我在您的解释中遗漏了某些内容，否则您不需要AttributeMask。如果最终的列名称只是原始列名称，然后是Kind值，那么您可以使用：

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;

见SQL Fiddle with Demo。这给出了结果：

| 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 |

为PIVOT源排序数据

1 个答案: