用于在每个层次结构级别聚合成本的 SQL Server CTE

Question

我不知道如何编写一个 CTE，它将成本从事务表汇总到自联接表，以便在层次结构的每个级别为我提供总成本。我整理了一个非常简单的例子来说明这个问题。 以下是 DDL 和插入脚本，以便您可以重现该问题，如果您愿意帮助我的话：

CREATE TABLE [Items](
    [ItemId] [int] IDENTITY(1,1) NOT NULL,
    [ParentId] [int] NULL,
    [ItemName] [varchar](100) NOT NULL,
 CONSTRAINT [PK_Items] PRIMARY KEY CLUSTERED 
(
    [ItemId] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
) ON [PRIMARY]
GO
CREATE TABLE [Transactions](
    [TransactionId] [int] IDENTITY(1,1) NOT NULL,
    [ItemId] [int] NOT NULL,
    [Amount] [money] NOT NULL,
 CONSTRAINT [PK_Transactions] PRIMARY KEY CLUSTERED 
(
    [TransactionId] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
) ON [PRIMARY]
GO
SET IDENTITY_INSERT [Items] ON 
GO
INSERT [Items] ([ItemId], [ParentId], [ItemName]) VALUES (1, NULL, N'Warehouse')
GO
INSERT [Items] ([ItemId], [ParentId], [ItemName]) VALUES (3, 1, N'Bin 1')
GO
INSERT [Items] ([ItemId], [ParentId], [ItemName]) VALUES (4, 1, N'Bin 2')
GO
INSERT [Items] ([ItemId], [ParentId], [ItemName]) VALUES (5, 3, N'Item 1.1')
GO
INSERT [Items] ([ItemId], [ParentId], [ItemName]) VALUES (6, 3, N'Item 1.2')
GO
INSERT [Items] ([ItemId], [ParentId], [ItemName]) VALUES (7, 4, N'Item 2.1')
GO
INSERT [Items] ([ItemId], [ParentId], [ItemName]) VALUES (8, 4, N'Item 2.2')
GO
SET IDENTITY_INSERT [Items] OFF
GO
SET IDENTITY_INSERT [Transactions] ON 
GO
INSERT [Transactions] ([TransactionId], [ItemId], [Amount]) VALUES (1, 5, 10.0000)
GO
INSERT [Transactions] ([TransactionId], [ItemId], [Amount]) VALUES (2, 5, 10.0000)
GO
INSERT [Transactions] ([TransactionId], [ItemId], [Amount]) VALUES (3, 6, 10.0000)
GO
INSERT [Transactions] ([TransactionId], [ItemId], [Amount]) VALUES (4, 6, 10.0000)
GO
INSERT [Transactions] ([TransactionId], [ItemId], [Amount]) VALUES (5, 4, 10.0000)
GO
INSERT [Transactions] ([TransactionId], [ItemId], [Amount]) VALUES (6, 7, 10.0000)
GO
INSERT [Transactions] ([TransactionId], [ItemId], [Amount]) VALUES (7, 7, 10.0000)
GO
INSERT [Transactions] ([TransactionId], [ItemId], [Amount]) VALUES (8, 8, 10.0000)
GO
INSERT [Transactions] ([TransactionId], [ItemId], [Amount]) VALUES (9, 8, 10.0000)
GO
SET IDENTITY_INSERT [Transactions] OFF
GO
ALTER TABLE [Items]  WITH CHECK ADD  CONSTRAINT [FK_Items_Items] FOREIGN KEY([ParentId])
REFERENCES [Items] ([ItemId])
GO
ALTER TABLE [Items] CHECK CONSTRAINT [FK_Items_Items]
GO
ALTER TABLE [Transactions]  WITH CHECK ADD  CONSTRAINT [FK_Transactions_Items] FOREIGN KEY([ItemId])
REFERENCES [Items] ([ItemId])
GO
ALTER TABLE [Transactions] CHECK CONSTRAINT [FK_Transactions_Items]
GO

这是我一直在研究的 CTE：

With cteAggregateCost
as  
(  
    select i.itemId, i.ParentId, t.Amount
    from Items i join Transactions t on i.ItemId = t.ItemId
union all
    select i.itemId, i.ParentId, t.Amount
    from Items i join cteAggregateCost c on i.ItemId = c.ParentId
    join Transactions t on i.ItemId = t.ItemId
)
select i.ParentId, i.ItemId, i.ItemName, sum(Amount) As AggregateCost
from Items i left join cteAggregateCost c on i.ItemId = c.ItemId
group by i.ParentId, i.ItemId, i.ItemName

这是我得到的结果：

这是我希望得到的结果：

如您所见，除前两行外，所有行都正常工作，这对容器没有成本，只有它们包含的项目。

非常感谢您提供的任何指导！

Answer 1

你可以像这样使用recursive CTE

;WITH temp AS 
(
   SELECT i.*, sum(isnull(t.Amount,0)) AS Amount 
   FROM @Items i
   LEFT JOIN @Transactions t ON t.ItemId = i.ItemId
   GROUP BY i.ItemId, i.ParentId, i.ItemName
)
,cteAggregateCost
as  
(  
   select i.ItemId, i.ItemId AS RootId, i.Amount  
   from temp i     
union all
   select i.ItemId, c.RootId, i.Amount 
   from cteAggregateCost c
   INNER JOIN temp i ON i.ParentId = c.ItemId
)
select i.*, ca.TotalAmount
from  @Items i 
CROSS APPLY 
(
   SELECT Sum(cac.Amount) AS TotalAmount
   FROM cteAggregateCost cac WHERE i.ItemId = cac.RootId
) ca
OPTION (MAXRECURSION 0)

演示链接：http://rextester.com/XMK96314

Answer 2

谢谢你，TriV。你的回答太棒了！但是，我确实找到了Xi Jin on the SQL Server forum 发布的更简单的答案。这是他的解决方案：

    With cteAggregateCost
    as  
    (  
        select  i.itemId as rootid,i.itemid, i.ParentId
        from Items i 
        union all
        select rootid, i.itemId, i.ParentId
        from Items i join cteAggregateCost c on i.ParentId = c.ItemId
    )

    select a.parentid, a.ItemId , a.ItemName , sum(t.Amount) As AggregateCost
    from items a
    left join cteAggregateCost i on a.itemid = i.rootid 
    left join Transactions t on i.ItemId = t.ItemId
    group by a.parentid, a.ItemId, a.ItemName

当针对具有多层层次关系的更大数据集进行测试时，这两种解决方案都给出了相同的正确结果。对我来说，西林的回答比较容易理解。我只是不知道如何添加 rootID 技术来保留没有自己成本且只有子项目成本的项目的值。