sql-server - 如何替换一个功能性的(很多)OUTER APPLY (SELECT * FROM)

Question

适用于 Microsoft SQL Server 2008 R2。

问题是

如果我们有几十个 Outer Apply (30)，那么它们开始工作的速度会很慢。在 Outer Apply 中间，我有一些比简单选择更复杂的东西，一个 View 。

详情

我正在编写一种分配给表(在数据库中)的属性。通常，几个表，保存对表的属性(键，值)的引用。

伪结构如下所示:

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

INSERT INTO @Lot
OUTPUT INSERTED.*
VALUES ('Hello'), ('World')

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

INSERT INTO @Attribute
OUTPUT INSERTED.* VALUES 
(1, 'Foo1', 'Kind1'), (1, 'Foo2', 'Kind2'), 
(2, 'Bar1', 'Kind1'), (2, 'Bar2', 'Kind2'), (2, 'Bar3', 'Kind3')

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

我现在可以运行如下查询:

SELECT 
[l].[LotId]
  , [SomeText]
  , [Oa1].[AttributeId]
  , [Oa1].[LotId]
  , 'Kind1Val' = [Oa1].[Val]
  , [Oa1].[Kind]
  , [Oa2].[AttributeId]
  , [Oa2].[LotId]
  , 'Kind2Val' = [Oa2].[Val]
  , [Oa2].[Kind]
  , [Oa3].[AttributeId]
  , [Oa3].[LotId]
  , 'Kind3Val' = [Oa3].[Val]
  , [Oa3].[Kind]  
FROM @Lot AS l
OUTER APPLY(SELECT * FROM @Attribute AS la WHERE la.[LotId] = l.[LotId] AND la.[Kind] = 'Kind1') AS Oa1
OUTER APPLY(SELECT * FROM @Attribute AS la WHERE la.[LotId] = l.[LotId] AND la.[Kind] = 'Kind2') AS Oa2
OUTER APPLY(SELECT * FROM @Attribute AS la WHERE la.[LotId] = l.[LotId] AND la.[Kind] = 'Kind3') AS Oa3


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

获取属性值透视表的简单方法和没有属性的 Lot 行的结果，如 Kind3。我知道微软PIVOT它并不简单，不适合放在这里。

最后，什么会更快并且会产生相同的结果？

Answer 1

为了获得结果，您可以反透视 然后透视 数据。

您可以通过两种方式执行此操作。首先，您可以使用 UNPIVOT 和 PIVOT 函数:

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.attributeid) rn
        from @Lot l
        left join @Attribute a
            on l.LotId = a.LotId
    ) src
    unpivot
    (
        value
        for col in (attributeid, a_Lotid, val, kind)
    ) unpiv
) d
pivot
(
    max(value)
    for col in (attributeid_1, a_LotId_1, Val_1, Kind_1,
                attributeid_2, a_LotId_2, Val_2, Kind_2,
                attributeid_3, a_LotId_3, Val_3, Kind_3)
) piv

参见 SQL Fiddle with Demo .

或者从 SQL Server 2008+ 开始，您可以使用带有 VALUES 子句的CROSS APPLY 来逆透视数据:

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.attributeid) 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

参见 SQL Fiddle with Demo .

unpivot 过程为每个 LotID 和 SomeText 获取多个列，并将其转换为行，给出结果:

| LOTID | SOMETEXT |           COL | VALUE |
--------------------------------------------
|     1 |    Hello | attributeid_1 |     1 |
|     1 |    Hello |       LotId_1 |     1 |
|     1 |    Hello |       Value_1 |  Foo1 |
|     1 |    Hello |        Kind_1 | Kind1 |
|     1 |    Hello | attributeid_2 |     2 |

我向内部子查询添加了一个 row_number() 以用于创建新的列名称以进行透视。创建名称后，可以将数据透视表应用于新列以给出最终结果

这也可以使用动态 SQL 来完成:

DECLARE @cols AS NVARCHAR(MAX),
    @query  AS NVARCHAR(MAX)

select @cols = STUFF((SELECT ',' + QUOTENAME(col+'_'+rn) 
                    from 
                    (
                      select 
                        cast(ROW_NUMBER() over(partition by l.lotid order by a.attributeid) as varchar(10)) rn
                      from Lot l
                      left join Attribute a
                          on l.LotId = a.LotId
                    ) t
                    cross apply (values ('attributeid', 1),
                                 ('LotId', 2), 
                                 ('Value', 3), 
                                 ('Kind', 4)) c (col, so)
                    group by col, rn, so
                    order by rn, so
            FOR XML PATH(''), TYPE
            ).value('.', 'NVARCHAR(MAX)') 
        ,1,1,'')


set @query = 'SELECT LotId,
                    SomeText,' + @cols + ' 
             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.attributeid) 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)
            ) x
            pivot 
            (
                max(value)
                for col in (' + @cols + ')
            ) p '

execute(@query)

参见 SQL Fiddle with Demo

所有三个版本都会给出相同的结果:

| 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 |             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 |