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[英]how to find the last level child nodes in tree kind structure in database using SQL
[英]SQL - How to find a value in a tree level data structure
我有两个SQL Server
表:
invoice
) invoice_relation
) invoice
表存储具有事务作品集的所有发票记录。
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:
因此,目标是在invoice
表下找到“ folio
”,给出一个invoicenumber
和folio
但folio
有时不会是invoice
的folio
,所以我需要搜索所有树关系才能找到如果任何发票与发票编号匹配,但是folio
是该关系的一部分。
例如,我必须找到对开的发票编号:
在我的查询中,我需要首先通过作品集查找,因为它是我的invoice
表主键。 然后,将尝试匹配A1122
与不匹配的D1122
,因此我必须搜索所有树结构以查找是否存在A1122
。 结果是在对开页1000
中找到了发票A1122
。
有关如何做到这一点的任何线索?
这是一个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
我仍然不确定你真正想要什么,我写了类似JamieD77的东西,找到顶级父母,然后走回树下但是你得到的孩子和granchildren与A1122没有直接关系.....
这是一种在树上走来走去的方式,让所有直接与发票人相关的孩子和父母归来
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
但根据你究竟在寻找什么,你可能会得到几个不想要的堂兄......
--------------这是一个找到顶级父母并获得整棵树的方法
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
你的模型一开始就被打破了。 parentinvoice应该在发票表中。 它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以为空的外键。 该字段(父发票字段)为空时,表示它是主发票。 任何有父母的行都是一张发票。
如果要在树级结构中查找值,可以将初始sql查询包装到“SELECT(.....)”语句中(创建自己的自定义可选表),过滤掉所需的内容。 如果您有任何疑问,请告诉我!
我对你的实际要求有点不清楚,所以我认为表值函数可能适合这里。 我添加了一些可选项,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)。 它们提供许多功能:表示顺序,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Folio的PATH
Select * from [dbo].[udf_SomeFunction](NULL,'1003')
将Folio限制为发票
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:
这当然可能比你看到的更多,而且很有可能我完全误解了你的要求。 也就是说,作为一个TVF,您可以使用其他WHERE和/或ORDER子句进行扩展,甚至可以合并到CROSS APPLY中。
这使用了使用hierarchyid
的方法,首先为每一行生成hierarchyid,然后选择folio为1003的行,然后查找具有'A1122'的invoicenumber的所有祖先。 这不是很有效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'
编辑了由@jj32突出显示的案例,您希望在其中找到对开页1003所在的层次结构中的根元素,然后查找该根的任何后代,其发票号为“A1122”。 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'
这很棘手,因为你有一个单独的关系表,根本发票不在其中。
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
这是一种尝试,首先使关系变得平坦,以便你可以向任何方向旅行。 然后它执行递归CTE来完成各个级别:
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应该在发票表中。 如果您知道发票之间的最大分隔级别,也可以在没有递归CTE的情况下完成此操作。
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