Select 带有版本控制的行的最终状态

Select final state of a row with versioning

我有一个table这样的

ID  Value1  Value2  value3  Versioning
1   sport   tennis    2         1
1   NULL    NULL      4         2
1   NULL    football NULL       3
1   game    NULL     NULL       4

这实际上是一个从一个数据库到另一个数据库的自定义复制table。逻辑如下: 您复制的第一行 (versioning=1) 带有它的所有字段。然后,每次对原始 table 进行任何更新时,只会复制更改的值,而不是整个 table。所以在 4 个版本之后,我们最终会像上面那样。我需要做的是创建一个可以读取此 table 和 return 只有一行实际上是最后状态的查询。 使用我们的示例 table 我想要的结果将是

ID Value1 Value2      Value3
1   game   football     4

在第一个版本中解释 Value1 int 的结果,我在第二个和第三个版本中有 'sport',我们没有任何更改,在第四个版本中它已更新为 'game'。分别对于其他值,我们有 tennis -> No change -> football -> No change 而对于值 3,我们有 2 -> 4-> No change -> 每个都没有变化 -> 代表一个版本。

这在 SQL 服务器中相当棘手,因为它不支持 window 函数上的 ignore nulls 选项。您可以使用重复的 applys,每列一个:

select t.id, t1.value1, t2.value2, t3.value3
from (values (1)) t(id) outer apply
     (select top (1) t2.value1
      from yourtable t1
      where t1.id = t.id and t1.value1 is not null
      order by t1.versioning desc
     ) t1 outer apply 
     (select top (1) t2.value1
      from yourtable t2
      where t2.id = t.id and t2.value1 is not null
      order by t2.versioning desc
     ) t2 outer apply 
     (select top (1) t3.value1
      from yourtable t3
      where t3.id = t.id and t3.value1 is not null
      order by t3.versioning desc
     ) t3;

With a CTE that returns the [Versioning] for the latest non null value of each of the columns [ValueX] 然后加入到 table :

with cte as (
  select [ID],
    max(case when [Value1] is not null then [Versioning] end) v1,
    max(case when [Value2] is not null then [Versioning] end) v2,
    max(case when [Value3] is not null then [Versioning] end) v3
  from tablename
  group by [ID]
)
select c.[ID], t1.[Value1], t2.[Value2], t3.[Value3]
from cte c
inner join tablename t1 on t1.[ID] = c.[ID] and t1.[Versioning] = c.v1
inner join tablename t2 on t2.[ID] = c.[ID] and t2.[Versioning] = c.v2
inner join tablename t3 on t3.[ID] = c.[ID] and t3.[Versioning] = c.v3

参见demo
结果:

> ID | Value1 | Value2   | Value3
> -: | :----- | :------- | :-----
>  1 | game   | football | 4     

您可以通过将您的版本控制和您的值合并到一个二进制列中,然后选择最大值来做到这一点。最短形式的查询将是:

SELECT  t.ID,
        Value1 = CONVERT(VARCHAR(50), SUBSTRING(MAX(CONVERT(BINARY(4), t.Versioning) 
                    + CONVERT(VARBINARY(50), t.Value1)), 5, 50)),
        Value2 = CONVERT(VARCHAR(50), SUBSTRING(MAX(CONVERT(BINARY(4), t.Versioning) 
                    + CONVERT(VARBINARY(50), t.Value2)), 5, 50)),
        Value3 = CONVERT(INT, SUBSTRING(MAX(CONVERT(BINARY(4), t.Versioning) 
                    + CONVERT(VARBINARY(50), t.Value3)), 5, 50))
FROM    YourTable AS t
GROUP BY ID;

为了解释发生了什么,我将只关注值 3,以及缩减示例数据。

该过程的第一步只是将排序列和值列组合成一个二进制值:

SELECT  *,
        BinaryValue3 = CONVERT(BINARY(2), t.Versioning)  + CONVERT(BINARY(2), t.Value3)     
FROM    (VALUES (1, 2, 1), (1, 4, 2), (1, NULL, 3)) AS t (ID, Value3, Versioning)

给出:

ID      Value3  Versioning      BinaryValue3
--------------------------------------
1       2           1           0x00010002
1       4           2           0x00020004
1       NULL        3           NULL

然后我们取二进制值的最大值。这取决于两件事:

  1. 连接 NULL 将产生 NULL,因此非空记录只有一个二进制值
  2. 由于二进制值将从左到右排序,MAX 函数将始终选择版本号最高的二进制值

然后一旦我们有了最大二进制值 (0x00020004),这只是提取右侧并将其转换回原始数据类型的情况。

完整的工作演示

DECLARE @T TABLE 
(
    ID INT NOT NULL, 
    Value1  VARCHAR(50), 
    Value2 VARCHAR(50), 
    value3 INT, 
    Versioning INT NOT NULL,
    PRIMARY KEY (ID, Versioning)
);
INSERT @T (ID, Value1, Value2, Value3, Versioning)
VALUES
    (1, 'sport', 'tennis', 2, 1),
    (1, NULL, NULL, 4, 2),
    (1, NULL, 'football', NULL, 3),
    (1, 'game', NULL, NULL, 4);

SELECT  t.ID,
        Value1 = CONVERT(VARCHAR(50), SUBSTRING(MAX(CONVERT(BINARY(4), t.Versioning) 
                    + CONVERT(VARBINARY(50), t.Value1)), 5, 50)),
        Value2 = CONVERT(VARCHAR(50), SUBSTRING(MAX(CONVERT(BINARY(4), t.Versioning) 
                    + CONVERT(VARBINARY(50), t.Value2)), 5, 50)),
        Value3 = CONVERT(INT, SUBSTRING(MAX(CONVERT(BINARY(4), t.Versioning) 
                    + CONVERT(VARBINARY(50), t.Value3)), 5, 50))
FROM    @T AS t
GROUP BY ID;

您还可以将此方法与 window 函数一起使用,将最后一个非空值添加到每一行,因此如果您想填充所有空值,您可以使用最后一个非空值:

DECLARE @T TABLE 
(
    ID INT NOT NULL, 
    Value1  VARCHAR(50), 
    Value2 VARCHAR(50), 
    value3 INT, 
    Versioning INT NOT NULL,
    PRIMARY KEY (ID, Versioning)
);
INSERT @T (ID, Value1, Value2, Value3, Versioning)
VALUES
    (1, 'sport', 'tennis', 2, 1),
    (1, NULL, NULL, 4, 2),
    (1, NULL, 'football', NULL, 3),
    (1, 'game', NULL, NULL, 4);

SELECT  t.ID,
        ActualValue1 = t.Value1,
        ActualValue2 = t.Value2,
        ActualValue3 = t.Value3,
        LastNonNUllValue1 = CONVERT(VARCHAR(50), SUBSTRING(MAX(Value1Bin) OVER(PARTITION BY t.ID ORDER BY t.Versioning), 5, 50)),
        LastNonNUllValue2 = CONVERT(VARCHAR(50), SUBSTRING(MAX(Value2Bin) OVER(PARTITION BY t.ID ORDER BY t.Versioning), 5, 50)),
        LastNonNUllValue3 = CONVERT(INT, SUBSTRING(MAX(Value3Bin) OVER(PARTITION BY t.ID ORDER BY t.Versioning), 5, 50)),
        t.Versioning
FROM    @T AS t
        CROSS APPLY
        (   SELECT  Value1Bin = CONVERT(BINARY(4), t.Versioning) + CONVERT(VARBINARY(50), t.Value1),
                    Value2Bin = CONVERT(BINARY(4), t.Versioning) + CONVERT(VARBINARY(50), t.Value2),
                    Value3Bin = CONVERT(BINARY(4), t.Versioning) + CONVERT(VARBINARY(50), t.Value3)
        ) AS b
ORDER BY t.Versioning;

给出:

ID  ActualValue1    ActualValue2    ActualValue3    LastNonNUllValue1   LastNonNUllValue2   LastNonNUllValue3   Versioning
------------------------------------------------------------------------------------------------------------------------------
1   sport           tennis          2               sport               tennis                  2                   1
1   NULL            NULL            4               sport               tennis                  4                   2
1   NULL            football        NULL            sport               football                4                   3
1   game            NULL            NULL            game                football                4                   4

有关更多阅读,请参阅 Itzik Ben-Gan's The Last non NULL Puzzle

还有一个选择。

这里我们unpivot你的数据,然后pivot

例子

Select *
 From  (
        Select Top 1 with ties 
               A.ID
              ,B.*
         From  YourTable A
         Cross Apply ( values ('Value1',Value1)
                             ,('Value2',Value2)
                             ,('Value3',convert(varchar(50),Value3))
                     ) B(Item,Value)
          Where Value is not null
          Order By row_number() over (partition by id,item order by versioning desc)
       ) pvt
 Pivot (max(value) for item in ([Value1],[Value2],[Value3]) ) pvt

Returns

ID  Value1  Value2      Value3
1   game    football    4

假设您的 table 是 'tablename' 并且以下代码提供最新值并且可以扩展到任何其他列

select 
(SELECT TOP 1 Value1 FROM tablename WHERE Value1 IS NOT NULL ORDER BY Versioning desc) Value1,
(SELECT TOP 1 Value2 FROM tablename WHERE Value2 IS NOT NULL ORDER BY Versioning desc) Value2,
(SELECT TOP 1 Value3 FROM tablename WHERE Value3 IS NOT NULL ORDER BY Versioning desc) Value3