SQL - 如何在树级数据结构中查找值

Question

我有两个SQL Server 表：

• 发票 (invoice)
• 发票关系 (invoice_relation)

invoice 表存储所有带有交易对账单的发票记录。

invoice_relation 表存储发票之间的任何关系。

这是发票如何相互关联的示例：

所以目标是在给定invoicenumber 和folio 的invoice 表下找到“folio”，但folio 有时不会是invoice 所具有的folio ，所以我需要对所有树关系进行搜索，以查找是否有任何发票与发票编号匹配，而且 folio 是关系的一部分。

例如，我必须找到对开并匹配以下发票编号：

• 对开页：1003
• 发票编号：A1122

在我的查询中，我需要先按 folio 查找，因为它是我的 invoice 表主键。然后，将尝试将A1122 与不匹配的D1122 匹配，因此我必须搜索所有树结构以查找是否有A1122。结果将是发票 A1122 在文件夹 1000 中找到。

关于如何做到这一点的任何线索？

以下是如何使用数据创建上述示例表的脚本：

CREATE TABLE [dbo].[invoice](
    [folio] [int] NOT NULL,
    [invoicenumber] [nvarchar](20) NOT NULL,
    [isactive] [bit] NOT NULL,
 CONSTRAINT [PK_invoice] PRIMARY KEY CLUSTERED 
(
    [folio] 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 [dbo].[invoice_relation](
    [relationid] [int] NOT NULL,
    [invoice] [nvarchar](20) NOT NULL,
    [parentinvoice] [nvarchar](20) NOT NULL,
 CONSTRAINT [PK_invoice_relation_1] PRIMARY KEY CLUSTERED 
(
    [relationid] 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
INSERT [dbo].[invoice] ([folio], [invoicenumber], [isactive]) VALUES (1000, N'A1122', 1)
GO
INSERT [dbo].[invoice] ([folio], [invoicenumber], [isactive]) VALUES (1001, N'B1122', 1)
GO
INSERT [dbo].[invoice] ([folio], [invoicenumber], [isactive]) VALUES (1002, N'C1122', 1)
GO
INSERT [dbo].[invoice] ([folio], [invoicenumber], [isactive]) VALUES (1003, N'D1122', 1)
GO
INSERT [dbo].[invoice] ([folio], [invoicenumber], [isactive]) VALUES (1004, N'F1122', 1)
GO
INSERT [dbo].[invoice] ([folio], [invoicenumber], [isactive]) VALUES (1005, N'G1122', 1)
GO
INSERT [dbo].[invoice_relation] ([relationid], [invoice], [parentinvoice]) VALUES (1, N'A1122', N'B1122')
GO
INSERT [dbo].[invoice_relation] ([relationid], [invoice], [parentinvoice]) VALUES (2, N'C1122', N'A1122')
GO
INSERT [dbo].[invoice_relation] ([relationid], [invoice], [parentinvoice]) VALUES (3, N'D1122', N'A1122')
GO
INSERT [dbo].[invoice_relation] ([relationid], [invoice], [parentinvoice]) VALUES (4, N'F1122', N'B1122')
GO
INSERT [dbo].[invoice_relation] ([relationid], [invoice], [parentinvoice]) VALUES (5, N'G1122', N'F1122')
GO

Answer 1

我仍然不确定你真正想要什么，我写了一些类似于 JamieD77 的东西，即找到顶级父母，然后从树上走下来，但是你得到的孩子和孙子与 A1122 没有直接关系...... .

这是一种在树上走来走去并返回与发票编号直接相关的所有孩子和父母的方法

DECLARE @InvoiceNumber NVARCHAR(20) = 'A1122'
DECLARE @Folio INT = 1003

;WITH cteFindParents AS (
    SELECT
       i.folio
       ,i.invoicenumber
       ,CAST(NULL AS NVARCHAR(20)) as ChildInvoiceNumber
       ,CAST(NULL AS NVARCHAR(20)) as ParentInvoiceNumber
       ,0 as Level
    FROM
       dbo.invoice i
    WHERE
       i.invoicenumber = @InvoiceNumber

    UNION ALL

    SELECT
       i.folio
       ,i.invoicenumber
       ,c.invoicenumber as ChildInvoiceNumber
       ,i.invoicenumber as ParentInvoiceNumber
       ,c.Level - 1 as Level
    FROM
       cteFindParents c
       INNER JOIN dbo.invoice_relation r
       ON c.invoicenumber = r.invoice
       INNER JOIN dbo.invoice i
       ON r.parentinvoice = i.invoicenumber
)

, cteFindChildren as (
    SELECT *
    FROM
       cteFindParents

    UNION ALL

    SELECT
       i.folio
       ,i.invoicenumber
       ,i.invoicenumber AS ChildInvoiceNumber
       ,c.invoicenumber AS ParentInvoiceNumber
       ,Level + 1 as Level
    FROM
       cteFindChildren c
       INNER JOIN dbo.invoice_relation r
       ON c.invoicenumber = r.parentinvoice
       INNER JOIN dbo.invoice i
       ON r.invoice = i.invoicenumber
    WHERE
       c.Level = 0
)

SELECT *
FROM
    cteFindChildren

但是根据您要寻找的具体内容，您实际上可能会得到几个不想要的表亲.....

-------------这是一种找到顶级父并获取整棵树的方法

DECLARE @InvoiceNumber NVARCHAR(20) = 'A1122'
DECLARE @Folio INT = 1003

;WITH cteFindParents AS (
    SELECT
       i.folio
       ,i.invoicenumber
       ,CAST(NULL AS NVARCHAR(20)) as ChildInvoiceNumber
       ,0 as Level
    FROM
       dbo.invoice i
    WHERE
       i.invoicenumber = @InvoiceNumber

    UNION ALL

    SELECT
       i.folio
       ,i.invoicenumber
       ,c.invoicenumber as ChildInvoiceNumber
       ,c.Level + 1 as Level
    FROM
       cteFindParents c
       INNER JOIN dbo.invoice_relation r
       ON c.invoicenumber = r.invoice
       INNER JOIN dbo.invoice i
       ON r.parentinvoice = i.invoicenumber

)

, cteGetTopParent AS  (
    SELECT *, ROW_NUMBER() OVER (PARTITION BY 1 ORDER BY LEVEL DESC) as RowNum
    FROM
       cteFindParents
)

, cteGetWholeTree AS (
    SELECT
       p.folio
       ,p.invoicenumber
       ,p.invoicenumber as TopParent
       ,p.invoicenumber as Parent
       ,CAST(p.invoicenumber AS NVARCHAR(1000)) as Hierarchy
       ,0 as Level
    FROM
       cteGetTopParent p
    WHERE
       RowNum = 1

    UNION ALL

    SELECT
       i.folio
       ,i.invoicenumber
       ,c.TopParent
       ,c.invoicenumber AS Parent
       ,CAST(c.TopParent + '|' + (CASE WHEN Level > 0 THEN c.invoicenumber + '|' ELSE '' END) + i.invoicenumber  AS NVARCHAR(1000)) as Hierarchy
       ,Level + 1 as Level
    FROM
       cteGetWholeTree c
       INNER JOIN dbo.invoice_relation r
       ON c.invoicenumber = r.parentinvoice
       INNER JOIN dbo.invoice i
       ON r.invoice = i.invoicenumber
)

SELECT *
FROM
    cteGetWholeTree

Answer 2

你的模型一开始就坏了。 parentinvoice 应该在 invoice 表中。这是一个递归数据库模型......所以使表模式递归。在引用它自己的表的列中有一个可为空的外键。任何时候该字段（父发票字段）为空都表示它是主发票。任何有父级的行都是一张发票。

当您想在树级结构中查找值时，您可以将初始 sql 查询包装到“SELECT(.....)”语句（创建您自己的自定义可选择表）中，该语句会过滤掉您想要的内容。如果您有任何问题，请告诉我！

Answer 3

我对您的实际要求有点不清楚，所以我认为表值函数在这里可能是合适的。我添加了一些可选项目，如果不需要，它们很容易删除（即 TITLE、Nesting、TopInvoice、TopFolio）。此外，您可能会注意到范围键 (R1/R2)。它们有许多功能：表示序列、选择标准、父/叶指示符，也许最重要的是非递归聚合。

返回整个层次结构

Select * from [dbo].[udf_SomeFunction](NULL,NULL)     

---

退回发票及其所有后代

Select * from [dbo].[udf_SomeFunction]('A1122',NULL) 

---

返回作品集的 PATH

Select * from [dbo].[udf_SomeFunction](NULL,'1003') 

---

将 Folio Limited 退回发票

Select * from [dbo].[udf_SomeFunction]('A1122','1003')

以下代码需要 SQL 2012+

CREATE FUNCTION [dbo].[udf_SomeFunction](@Invoice nvarchar(25),@Folio nvarchar(25))
Returns Table
As  
Return (
with cteBld as (
      Select Seq  = cast(1000+Row_Number() over (Order By Invoice) as nvarchar(500)),I.Invoice,I.ParentInvoice,Lvl=1,Title = I.Invoice,F.Folio
        From (
              Select Distinct 
                     Invoice=ParentInvoice
                    ,ParentInvoice=cast(NULL as nvarchar(20)) 
              From   [Invoice_Relation] 
              Where  @Invoice is NULL and ParentInvoice Not In (Select Invoice from [Invoice_Relation])
              Union  All
              Select Invoice
                    ,ParentInvoice 
              From   [Invoice_Relation] 
              Where  Invoice=@Invoice
             ) I
        Join Invoice F on I.Invoice=F.InvoiceNumber
      Union  All
      Select Seq  = cast(concat(A.Seq,'.',1000+Row_Number() over (Order by I.Invoice)) as nvarchar(500))
            ,I.Invoice
            ,I.ParentInvoice
            ,A.Lvl+1
            ,I.Invoice,F.folio
      From   [Invoice_Relation] I
      Join   cteBld  A on I.ParentInvoice = A.Invoice 
      Join   Invoice F on I.Invoice=F.InvoiceNumber )
     ,cteR1  as (Select Seq,Invoice,Folio,R1=Row_Number() over (Order By Seq) From cteBld)
     ,cteR2  as (Select A.Seq,A.Invoice,R2=Max(B.R1) From cteR1 A Join cteR1 B on (B.Seq like A.Seq+'%') Group By A.Seq,A.Invoice )

Select Top 100 Percent 
       B.R1
      ,C.R2
      ,A.Invoice
      ,A.ParentInvoice
      ,A.Lvl
      ,Title = Replicate('|-----',A.Lvl-1)+A.Title    -- Optional: Added for Readability
      ,A.Folio
      ,TopInvoice  = First_Value(A.Invoice) over (Order By R1) 
      ,TopFolio    = First_Value(A.Folio)   over (Order By R1) 
 From  cteBld A 
 Join  cteR1  B on A.Invoice=B.Invoice 
 Join  cteR2  C on A.Invoice=C.Invoice 
 Where (@Folio is NULL)
    or (@Folio is Not NULL and (Select R1 from cteR1 Where Folio=@Folio) between R1 and R2)
 Order By R1
)

最后的想法：

这当然可能超出您的预期，而且我很可能完全误解了您的要求。也就是说，作为 TVF，您可以使用额外的 WHERE 和/或 ORDER 子句进行扩展，甚至可以合并到 CROSS APPLY 中。

Answer 4

这使用了一种使用hierarchyid 的方法，首先为每一行生成hierarchyid，然后选择folio 为1003 的行，然后查找所有具有发票编号“A1122”的祖先。效率不是很高，但可能会给你一些不同的想法：

;WITH
Allfolios
AS
(
    Select i.folio, i.InvoiceNumber,
          hierarchyid::Parse('/' + 
               CAST(ROW_NUMBER() 
                        OVER (ORDER BY InvoiceNumber) AS VARCHAR(30)
                   ) + '/') AS hierarchy, 1 as level
    from invoice i
    WHERE NOT EXISTS 
        (SELECT * FROM invoice_relation ir WHERE ir.invoice = i. invoicenumber)
    UNION ALL
    SELECT i.folio, i.invoiceNumber,
          hierarchyid::Parse(CAST(a.hierarchy as VARCHAR(30)) + 
                             CAST(ROW_NUMBER() 
                                   OVER (ORDER BY a.InvoiceNumber) 
                               AS VARCHAR(30)) + '/') AS hierarchy, 
          level + 1
    FROM Allfolios A
    INNER JOIN invoice_relation ir
        on a.InvoiceNumber = ir.ParentInvoice
    INNER JOIN invoice i
        on ir.Invoice = i.invoicenumber
),
Ancestors
AS
(
    SELECT folio, invoiceNumber, hierarchy, hierarchy.GetAncestor(1) as AncestorId
    from Allfolios
    WHERE folio = 1003
    UNION ALL
    SELECT af.folio, af.invoiceNumber, af.hierarchy, af.hierarchy.GetAncestor(1)
      FROM Allfolios AF
      INNER JOIN 
            Ancestors a ON Af.hierarchy= a.AncestorId
)
SELECT *
FROM Ancestors
WHERE InvoiceNumber = 'A1122'

针对@jj32 突出显示的情况进行了编辑，您希望在其中查找对开 1003 所在的层次结构中的根元素，然后查找发票编号为“A1122”的该根的任何后代。见下文：

;WITH
Allfolios -- Convert all rows to a hierarchy
AS
(
    Select i.folio, i.InvoiceNumber,
          hierarchyid::Parse('/' + 
               CAST(ROW_NUMBER() 
                        OVER (ORDER BY InvoiceNumber) AS VARCHAR(30)
                   ) + '/') AS hierarchy, 1 as level
    from invoice i
    WHERE NOT EXISTS 
        (SELECT * FROM invoice_relation ir WHERE ir.invoice = i. invoicenumber)
    UNION ALL
    SELECT i.folio, i.invoiceNumber,
          hierarchyid::Parse(CAST(a.hierarchy as VARCHAR(30)) + 
                             CAST(ROW_NUMBER() 
                                   OVER (ORDER BY a.InvoiceNumber) 
                               AS VARCHAR(30)) + '/') AS hierarchy, 
          level + 1
    FROM Allfolios A
    INNER JOIN invoice_relation ir
        on a.InvoiceNumber = ir.ParentInvoice
    INNER JOIN invoice i
        on ir.Invoice = i.invoicenumber
),
Root -- Find Root
AS
(
    SELECT *
    FROM AllFolios AF
    WHERE Level = 1 AND 
    (SELECT hierarchy.IsDescendantOf(AF.hierarchy)  from AllFolios AF2 WHERE folio = 1003) = 1
)
-- Find all descendants of the root element which have an invoicenumber = 'A1122'
SELECT *
FROM ALLFolios
WHERE hierarchy.IsDescendantOf((SELECT TOP 1 hierarchy FROM Root)) = 1 AND
invoicenumber = 'A1122'

Answer 5

这很棘手，因为您有一个单独的关系表并且根发票不在其中。

DECLARE @folio INT = 1003,
        @invoice NVARCHAR(20) = 'A1122'


-- find highest level of relationship
;WITH cte  AS (
    SELECT  i.folio,
            i.invoicenumber,
            ir.parentinvoice,
            0 AS [level]
    FROM    invoice i
            LEFT JOIN invoice_relation ir  ON ir.invoice = i.invoicenumber
    WHERE   i.folio = @folio
    UNION ALL 
    SELECT  i.folio,
            i.invoicenumber,
            ir.parentinvoice,
            [level] + 1
    FROM    invoice i
            JOIN invoice_relation ir  ON ir.invoice = i.invoicenumber
            JOIN cte r ON r.parentinvoice = i.invoicenumber
),
-- make sure you get the root folio
rootCte AS (
    SELECT  COALESCE(oa.folio, c.folio) AS rootFolio 
    FROM    (SELECT     *,
                        ROW_NUMBER() OVER (ORDER BY [level] DESC) Rn
                FROM    cte ) c
             OUTER APPLY (SELECT folio FROM invoice i WHERE i.invoicenumber = c.parentinvoice) oa
    WHERE    c.Rn = 1
),
-- get all children of root folio
fullTree AS (
    SELECT  i.folio,
            i.invoicenumber
    FROM    rootCte r
            JOIN invoice i ON r.rootFolio = i.folio
    UNION ALL 
    SELECT  i.folio,
            i.invoicenumber
    FROM    fullTree ft
            JOIN invoice_relation ir ON ir.parentinvoice = ft.invoicenumber
            JOIN invoice i ON ir.invoice = i.invoicenumber
)
-- search for invoice
SELECT  * 
FROM    fullTree
WHERE   invoicenumber = @invoice

Answer 6

这是一种首先拉平关系的尝试，以便您可以向任何方向移动。然后它执行递归 CTE 以通过各个级别：

WITH invoicerelation AS
(
select relationid, invoice, parentinvoice AS relatedinvoice
from invoice_relation
union
select relationid, parentinvoice AS invoice, invoice AS relatedinvoice 
from invoice_relation
),

cteLevels AS
(
select 0 AS relationid, invoice.folio, 
   invoicenumber AS invoice, invoicenumber AS relatedinvoice, 
   0 AS Level
from invoice 
UNION ALL
select invoicerelation.relationid, invoice.folio,
    invoicerelation.invoice, cteLevels.relatedinvoice, 
    Level + 1 AS Level
from invoice INNER JOIN
   invoicerelation ON invoice.invoicenumber = invoicerelation.invoice INNER JOIN
   cteLevels ON invoicerelation.relatedinvoice = cteLevels.invoice
      and (ctelevels.relationid <> invoicerelation.relationid)
)

SELECT cteLevels.folio, relatedinvoice, invoice.folio AS invoicefolio, cteLevels.level
from cteLevels INNER JOIN
    invoice ON cteLevels.relatedinvoice = invoice.invoicenumber
WHERE cteLevels.folio = 1003 AND cteLevels.relatedinvoice = 'a1122'

我同意 SwampDev 的评论，即 parentinvoice 确实应该在 invoice 表中。如果您知道发票之间的最大分隔级别数，也可以在没有递归 CTE 的情况下完成此操作。