Sql Select 使用 CTE 对递归数据进行排序

Question

我想要一个在 SQL 服务器

中使用 CTE 解决这个问题的方法

情况示例

Equation 0 = 0.25*Equation 1
Equation 1 = Equation 2 + Equation 3 + 0.5*Equation 5
Equation 2 = 15 + 40
Equation 3 = Equation 6 + Equation 7
Equation 4 = 10
Equation 5 = 10 + Equation 4
Equation 6 = 10 +5
Equation 7 = Equation 5 + Equation 2

tables 的结构是

元素Table

ID     | Name
-------|--------------
0      | Equation 0
1      | Equation 1
2      | Equation 2
3      | Equation 3
4      | Equation 4
5      | Equation 5
6      | Equation 6
7      | Equation 7
---------------------

table 包含每个等式的所有项目

等式Table

FK     |    Item   |    Type
-------|-----------|------------------
0      |    0.25   |    constant
0      |    *      |    Operator
0      |    1      |    Element
1      |    2      |    Element
1      |    +      |    Operator
1      |    3      |    Element
1      |    +      |    Operator
1      |    0.5    |    constant
1      |    *      |    Operator
1      |    5      |    Element
2      |    15     |    constant
2      |    +      |    Operator
2      |    40     |    constant
…      |           |      
…      |           | 
… etc  |           |        
------------------------------------

如果类型是元素，这意味着它是一个元素项目

是否有任何 sql 语句结果以正确的顺序我必须在不使用递归函数的情况下一个一个地计算这些方程，因为它在 SQL 中受到限制另一种方法是在没有任何要求的情况下计算最后一个方程，然后计算上面的方程，因为当我需要一个方程时，我发现它已经计算出来而无需递归方程

我需要 sql select 语句来生成以下命令

Equation 2
Equation 6
Equation 4
Equation 5
Equation 7
Equation 3
Equation 1
Equation 0

我是目测订购的，因为它很简单是否有任何 select 语句用于这样做或者用户必须手动订购？？？

...更新

1.具有完全工作的测试场景

要创建的脚本 tables

/****** Object:  Table [dbo].[Element]    Script Date: 26/03/2017 11:10:10 م ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE TABLE [dbo].[Element](
    [Id] [int] NOT NULL,
    [Name] [nvarchar](50) NOT NULL,
 CONSTRAINT [PK_Element] PRIMARY KEY CLUSTERED 
(
    [Id] 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
/****** Object:  Table [dbo].[Equation]    Script Date: 26/03/2017 11:10:10 م ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE TABLE [dbo].[Equation](
    [fk] [int] NOT NULL,
    [Item] [nvarchar](50) NOT NULL,
    [Type] [nvarchar](50) NOT NULL
) ON [PRIMARY]

GO
INSERT [dbo].[Element] ([Id], [Name]) VALUES (0, N'Equation 0')
INSERT [dbo].[Element] ([Id], [Name]) VALUES (1, N'Equation 1')
INSERT [dbo].[Element] ([Id], [Name]) VALUES (2, N'Equation 2')
INSERT [dbo].[Element] ([Id], [Name]) VALUES (3, N'Equation 3')
INSERT [dbo].[Element] ([Id], [Name]) VALUES (4, N'Equation 4')
INSERT [dbo].[Element] ([Id], [Name]) VALUES (5, N'Equation 5')
INSERT [dbo].[Element] ([Id], [Name]) VALUES (6, N'Equation 6')
INSERT [dbo].[Element] ([Id], [Name]) VALUES (7, N'Equation 7')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (0, N'0.25', N'constant')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (0, N'*', N'Operator')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (0, N'1', N'Element')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (1, N'2', N'Element')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (1, N'+', N'Operator')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (1, N'3', N'Element')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (1, N'+', N'Operator')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (1, N'0.5', N'constant')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (1, N'*', N'Operator')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (1, N'5', N'Element')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (2, N'15', N'constant')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (2, N'+', N'Operator')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (2, N'40', N'constant')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (3, N'6', N'Element')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (3, N'+', N'Operator')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (3, N'7', N'Element')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (4, N'10', N'Constant')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (5, N'10', N'Constant')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (5, N'+', N'Operator')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (5, N'4', N'Element')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (6, N'10', N'Constant')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (6, N'+', N'Operator')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (6, N'5', N'Constant')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (7, N'5', N'Element')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (7, N'+', N'Operator')
INSERT [dbo].[Equation] ([fk], [Item], [Type]) VALUES (7, N'2', N'Element')

*

@Gordon Linoff

...非常感谢您的关心，但我使用提供的脚本发现了一些我无法解决的错误，因为这是我第一次使用 CTE

第二个脚本给了我一些我无法解决的错误

第一个脚本也给我错误，我无法解决

Answer 1

哦，如果能表达成这样就好了：

with cte as (
      select e.fk, 1 as lev
      from equation e
      group by e.fk
      having sum(case when type = 'Element' then 1 else 0 end) = 0
      union all
      select e.fk, max(cte.lev) + 1
      from equation e left join
           cte
           on e.fk = cte.fk
      group by e.fk
      having count(*) = count(cte.fk)
     )

但这是不可能的。所以，我们必须考虑字符串操作（我认为）。这导致将依赖项放入一个字符串中，并重复地从字符串中删除元素。如果我有这个权利：

with eq as (
       select e.fk,
              stuff( (select ',' + e2.item
                      from equation e2
                      where e2.fk = e.fk and e2.type = 'Element'
                      order by e2.item
                      for xml path ('')
                     ), 1, 1, '') as elements 
       from (select distinct e.fk from equation e) e
      )
      select e.fk, '' as elements_found, 1 as lev
      from eq
      where elements = ''
      union all
      select eq.fk, substring(elements_found, charindex(',', elements_found + ',') + 1), 2 as lev
      from eq join
           cte
           on cte.elements_found like eq.fk + ',%' and eq.fk = cte.fk
      where eq.type = 'Element'
     )
select cte.fk, max(lev)
from cte
group by cte.fk
order by max(lev);

Answer 2

你可以这样试试：

WITH Related AS
(
    SELECT *
    FROM Equation AS eq
    LEFT JOIN Element AS e ON eq.[Type]='Element' AND eq.Item=CAST(e.Id AS VARCHAR(10))
    WHERE eq.[Type]='Element' 
)
,Dependecies AS
(
    SELECT e.*
          ,ISNULL(r.Name,'') AS DepName 
    FROM Element AS e
    LEFT JOIN Related AS r ON e.Id=r.fk
)
,recCTE AS
(
    SELECT 1 AS lvl,d.Id,d.Name,d.DepName
    FROM Dependecies AS d
    WHERE d.Name NOT IN(SELECT x.DepName FROM Dependecies AS x)

    UNION ALL

    SELECT r.lvl+1,d.Id,d.Name,d.DepName
    FROM recCTE AS r
    INNER JOIN Dependecies AS d ON r.DepName=d.Name

)
,Ranked AS
(
    SELECT Name
            ,DENSE_RANK() OVER(ORDER BY CASE WHEN DepName='' THEN 1000 ELSE lvl END DESC) AS Rnk
    FROM recCTE
)
SELECT Name,MIN(Rnk) AS Rnk
FROM Ranked
GROUP BY Name
ORDER BY Min(Rnk)

结果

Equation 2  1
Equation 4  1
Equation 6  1
Equation 5  2
Equation 7  3
Equation 3  4
Equation 1  5
Equation 0  6

说明

有一个 CTE 列表：

第一个 CTE 会将元素绑定到方程行，其中类型为 Element。
第二个将列出所有元素及其依赖项
第三个 CTE 是一个递归 CTE，从没有任何依赖关系的元素开始，沿着依赖路径向下移动
下一个 CTE 使用 DENSE_RANK() OVER() 来订购呼叫

最终SELECTreturns每个元素和最早需要它的时刻。

Sql Select 使用 CTE 对递归数据进行排序

Sql Select using CTE to order a recursive data

sql

sql-server

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recursive-query

common-table-expression

说明