Question

我对表中的记录进行了审核。有多列，每个记录状态，用于更改1列或更多列我需要返回一个审核结果，其中返回模式为：column（ id ， alias 或 name ），之前的值，新价值等问题在于每个新记录可能有多个列具有已更改的数据。与此同时，可审计列的数量为5，因此可以对其进行“硬编码”并更改验证。

那么是否可以以缩短的方式组合这样的查询，而不仅仅是使用UNIONS并为每列进行SELECT查询并检查更改？

假设有一个包含列的表：

id, datetime value, int value, varchar value.

如果我有2条记录，这样的数据更改如下：

id1, value1, value1, value1
id1, value2, value1, value2

然后我期待这样的审核结果：

id1, value1 as oldvalue, value2 as newvalue, column2name as columnname
id1, value1 as oldvalue, value2 as newvalue, column4name as columnname

Answer 1

如果我没有错过任何内容：

WITH ranked AS (
  SELECT
    ChangeDate,
    ColPK,
    Col1,
    Col2,
    Col3,
    Col4,
    Col5,
    OverallRank = ROW_NUMBER() OVER (PARTITION BY ColPK       ORDER BY ChangeDate),
    Col1Rank    = ROW_NUMBER() OVER (PARTITION BY ColPK, Col1 ORDER BY ChangeDate),
    Col2Rank    = ROW_NUMBER() OVER (PARTITION BY ColPK, Col2 ORDER BY ChangeDate),
    Col3Rank    = ROW_NUMBER() OVER (PARTITION BY ColPK, Col3 ORDER BY ChangeDate),
    Col4Rank    = ROW_NUMBER() OVER (PARTITION BY ColPK, Col4 ORDER BY ChangeDate),
    Col5Rank    = ROW_NUMBER() OVER (PARTITION BY ColPK, Col5 ORDER BY ChangeDate)
  FROM AuditTable
)
, ranked2 AS (
  SELECT
    ChangeDate,
    ColPK,
    Col1,
    Col2,
    Col3,
    Col4,
    Col5,
    Col1Group = RANK() OVER (PARTITION BY ColPK, Col1 ORDER BY OverallRank - Col1Rank),
    Col2Group = RANK() OVER (PARTITION BY ColPK, Col2 ORDER BY OverallRank - Col2Rank),
    Col3Group = RANK() OVER (PARTITION BY ColPK, Col3 ORDER BY OverallRank - Col3Rank),
    Col4Group = RANK() OVER (PARTITION BY ColPK, Col4 ORDER BY OverallRank - Col4Rank),
    Col5Group = RANK() OVER (PARTITION BY ColPK, Col5 ORDER BY OverallRank - Col5Rank),
    Col1Rank = ROW_NUMBER() OVER (PARTITION BY ColPK, Col1, OverallRank - Col1Rank ORDER BY ChangeDate),
    Col2Rank = ROW_NUMBER() OVER (PARTITION BY ColPK, Col2, OverallRank - Col2Rank ORDER BY ChangeDate),
    Col3Rank = ROW_NUMBER() OVER (PARTITION BY ColPK, Col3, OverallRank - Col3Rank ORDER BY ChangeDate),
    Col4Rank = ROW_NUMBER() OVER (PARTITION BY ColPK, Col4, OverallRank - Col4Rank ORDER BY ChangeDate),
    Col5Rank = ROW_NUMBER() OVER (PARTITION BY ColPK, Col5, OverallRank - Col5Rank ORDER BY ChangeDate)
  FROM ranked
),
unpivoted AS (
  SELECT
    r.ChangeTime,
    r.ColPK,
    x.ColName,
    ColRank = CASE x.Colname
      WHEN 'Col1' THEN Col1Group
      WHEN 'Col2' THEN Col2Group
      WHEN 'Col3' THEN Col3Group
      WHEN 'Col4' THEN Col4Group
      WHEN 'Col5' THEN Col5Group
    END,
    Value = CASE x.Colname
      WHEN 'Col1' THEN CONVERT(nvarchar(100), r.Col1)
      WHEN 'Col2' THEN CONVERT(nvarchar(100), r.Col2)
      WHEN 'Col3' THEN CONVERT(nvarchar(100), r.Col3)
      WHEN 'Col4' THEN CONVERT(nvarchar(100), r.Col4)
      WHEN 'Col5' THEN CONVERT(nvarchar(100), r.Col5)
    END
  FROM ranked2 r
    INNER JOIN (VALUES ('Col1'), ('Col2'), ('Col3'), ('Col4'), ('Col5')) x (ColName)
      ON x.ColName = 'Col1' AND Col1Rank = 1
      OR x.ColName = 'Col2' AND Col2Rank = 1
      OR x.ColName = 'Col3' AND Col3Rank = 1
      OR x.ColName = 'Col4' AND Col4Rank = 1
      OR x.ColName = 'Col5' AND Col5Rank = 1
)
SELECT
  new.ChangeTime,
  new.ColPK,
  new.ColName,
  old.Value AS OldValue,
  new.Value AS NewValue
FROM unpivoted new
  LEFT JOIN unpivoted old
    ON new.ColPK   = old.ColPK
   AND new.ColName = old.ColName
   AND new.ColRank = old.ColRank + 1

基本上，我们的想法是对相同值的连续组进行排名，并选择每个值的第一次出现。对于其值正在被审计的每个列都要执行此操作，并且在此过程中列不显示。之后，不相交的行集合自身，即对于每个PK和列名称，每一行都与其前一个行匹配（基于排名），以获得最终结果集的同一行中的旧值。

Answer 2

这是一个更简单的查询，可以产生相同的预期结果，并且更容易修改以容纳不同数量的列或更改列名称，因为唯一的区别是PK列+每个非单行CROSS APPLY中的-PK列。我必须添加一个ChangeDate列 - 没有它，就无法知道插入审计表的行的顺序。

WITH ColValues AS (
   SELECT
      Grp = Row_Number() OVER (
         PARTITION BY H.OrderID, U.ColName ORDER BY H.ChangeDate ASC, X.Which
      ) / 2,
      H.OrderID,
      H.ChangeDate,
      U.*,
      X.Which
   FROM
      dbo.OrderHistory H
      CROSS APPLY (VALUES
         ('DeliveryDate', Convert(varchar(1000), DeliveryDate, 121)),
         ('Quantity', Convert(varchar(1000), Quantity)),
         ('SpecialNotes', Convert(varchar(1000), SpecialNotes))
      ) U (ColName, Value)
      CROSS JOIN (VALUES (1), (2)) X (Which)
)
SELECT
   V.OrderID,
   V.ColName,
   DateChanged = Max(V.ChangeDate),
   OldValue = Max(F.Value),
   NewValue = Max(T.Value)
FROM
   ColValues V
   OUTER APPLY (SELECT V.ColName, V.Value WHERE V.Which = 2) F
   OUTER APPLY (SELECT V.ColName, V.Value WHERE V.Which = 1) T
GROUP BY
   V.OrderID,
   V.ColName,
   V.Grp
HAVING
   Count(*) = 2
   AND EXISTS (
      SELECT Max(F.Value)
      EXCEPT SELECT Max(T.Value)
   )
;

See a live demo of this query at SQL Fiddle

在SQL 2012中，使用LEAD或LAG分析函数可以更好地解决这个问题。我的查询中的CROSS JOIN和Row_Number通过复制每一行并将这些重复的行成对分配到它们自己的组（其中每个组有两行代表相邻的审计历史记录行）来模拟这一点。然后通过战略性地使用聚合，我们可以处理分组对以选择和比较它们的值。

另外，我最初使用UNPIVOT编写了查询，但是唉，它不会保留NULL - 在我看来，这是微软的严重疏忽。如果需要，开发人员可以很容易地添加删除NULL的条件，但是当希望保留NULL时，根本不能使用它UNPIVOT的方式。具有讽刺意味的是，使用CROSS APPLY到UNPIVOT时，生成的代码更加紧凑，缩短了2行 - 现在转换和取消转换只需一步而不是2步。

我的示例数据是：

ChangeDate              OrderID DeliveryDate            Quantity SpecialNotes
----------------------- ------- ----------------------- -------- ----------------------------------------------------
2013-03-01 11:28:00.000 1       2013-04-01 00:00:00.000 25       NULL
2013-03-01 11:56:00.000 1       2013-04-01 00:00:00.000 30       NULL
2013-03-05 10:18:00.000 1       2013-04-02 00:00:00.000 30       Customer called to ask for delivery date adjustment.
2013-03-01 11:37:00.000 2       2013-03-05 00:00:00.000 17       NULL

结果行集：

OrderID ColName      DateChanged             OldValue                NewValue
------- ------------ ----------------------- ----------------------- ---------------------------------------------------
1       DeliveryDate 2013-03-05 10:18:00.000 2013-04-01 00:00:00.000 2013-04-02 00:00:00.000
1       Quantity     2013-03-01 11:56:00.000 25                      30
1       SpecialNotes 2013-03-05 10:18:00.000 NULL                    Customer called to ask for delivery date adjustment.

注意：由于我的查询只有一个排名函数而且没有JOIN，因此即使在非常大的表中也会表现得非常好 - 或许比使用{{1没有支持索引的地方。审计表最好在JOIN上有一个聚簇索引。

SQL：返回每列记录的审核

2 个答案: