Sql Select using CTE to order a recursive data - sql

Question

Please i want a solution for this problem using CTE in SQL Server
Example for the situation
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
The structure of the tables is
The Element 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
---------------------
the table holds all items of each equation
The equation 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 | |
------------------------------------
if the type is element this means it is an element item
is there any sql statement result to the correct order that i must use to calculate these equations one by one without using recursive functions because it is limited in SQL
the alternative is to calculate the last equation without any requirements then calculate the upper ones as when i need an equation i find it calculated already without recursing the equations
i need the sql select statement to produce the following order
Equation 2
Equation 6
Equation 4
Equation 5
Equation 7
Equation 3
Equation 1
Equation 0
i ordered them by eye because it is simple example
is there any select statement is used to do so
or the user must order them manually ???
... Update
1. with a fully working test scenario
The Script to create 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
... Thanks a lot for your care but i found some errors using the provided script that i cant solve as it is the first time for me to use CTE
The second script gives me some errors that i cant solve
the first script gives me errors as well that i cant solve

Answer 1

You can try it like this:
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)
The result
Equation 2 1
Equation 4 1
Equation 6 1
Equation 5 2
Equation 7 3
Equation 3 4
Equation 1 5
Equation 0 6
Explanation
There is a list of CTEs:
The first CTE will bind the Elements to Equation rows, where the type is Element.
The second will list all Elements with their dependencies
The third CTE is a recursive CTE, starting with the element without any dependecies, working down the dependency path
The next CTE uses DENSE_RANK() OVER() to get the calls ordered
The final SELECT returns each element and the earliest moment it is needed.

Answer 2

Oh, it would be swell if we could express this as:
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)
)
But that is not possible. So, we have to think in terms of string manipulation (I think). This results in putting the dependencies in a string and repeatedly chopping elements off of the string. If I have this right:
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);