Hierarchy Tree
Itzik
============================================================================
if exists (select * from dbo.sysobjects where id =
object_id(N'[dbo].[RecursiveTable]') and OBJECTPROPERTY(id, N'IsUserTable')
= 1)
drop table [dbo].[RecursiveTable]
GO
CREATE TABLE [dbo].[RecursiveTable] (
[Parent] [int] NOT NULL ,
[Child] [int] NOT NULL ,
[Name] [varchar] (20) COLLATE Hebrew_CI_AS NOT NULL ,
[Price] [int] NOT NULL
) ON [PRIMARY]
GO
INSERT INTO RecursiveTable VALUES (0,1,'Dani',0)
INSERT INTO RecursiveTable VALUES (1,2,'Jonn',100)
INSERT INTO RecursiveTable VALUES (1,3,'Dorit',250)
INSERT INTO RecursiveTable VALUES (3,4,'Kimon',150)
INSERT INTO RecursiveTable VALUES (2,5,'Toni',400)
INSERT INTO RecursiveTable VALUES (3,6,'Popo',200)
============================================================================
=======
I need solution to build Hierarchy tree and calculate data
this data will be receive :
1 Dani 1100
2 Jonn 500
3 Toni 400
2 Dorit 600
3 Kimon 150
3 Popo 200
Thanks, Itzik
Jacco Schalkwijk
First, please don't post the same question twice, at least not at the same
day. People who answer questions on these newsgroups are volunteers, and
they might not have time to look at your question, specially if it is a
complex one.
I have two solutions for you, the first one is based on the table as you
provided it, and uses a table variable and some iteration to get the result.
The second one is based on Joe Celko's nested sets model. It takes to long
to explain that in a newspost, but Joe himself explaines in his article
here:
http://www.developersdex.com/gurus/articles/112.asp
You will notice that the nested sets model might be somewhat harder to
understand, but it definitly makes for far shorter code to get the results
you want.
CREATE TABLE [dbo].[RecursiveTable] (
[Parent] [int] NOT NULL ,
[Child] [int] NOT NULL ,
[Name] [varchar] (20) COLLATE Hebrew_CI_AS NOT NULL ,
[Price] [int] NOT NULL
) ON [PRIMARY]
GO
INSERT INTO RecursiveTable VALUES (0,1,'Dani',0)
INSERT INTO RecursiveTable VALUES (1,2,'Jonn',100)
INSERT INTO RecursiveTable VALUES (1,3,'Dorit',250)
INSERT INTO RecursiveTable VALUES (3,4,'Kimon',150)
INSERT INTO RecursiveTable VALUES (2,5,'Toni',400)
INSERT INTO RecursiveTable VALUES (3,6,'Popo',200)
DECLARE @result TABLE(level_no INT NOT NULL, [Name] [varchar] (20) COLLATE
Hebrew_CI_AS NOT NULL, total_price [int] NOT NULL, child int, parent int )
DECLARE @level_no INT
SET @level_no = 1
INSERT INTO @result(level_no, [name], total_price, child, parent)
SELECT @level_no, Name, price, child, parent FROM RecursiveTable WHERE
Parent = 0
WHILE 1=1
BEGIN
INSERT INTO @result(level_no, [name], total_price, child, parent)
SELECT @level_no +1, rt.Name, rt.price, rt.child, rt.parent FROM
RecursiveTable rt
INNER JOIN @result res
ON rt.parent = res.child
WHERE res.level_no = @level_no
IF @@ROWCOUNT = 0 BREAK
SET @level_no = @level_no +1
END
WHILE @level_no > 1
BEGIN
UPDATE r
SET total_price = total_price +
(SELECT SUM(total_price) FROM @result res
WHERE res.parent = r.child)
FROM @result r
WHERE level_no = @level_no - 1
SET @level_no = @level_no - 1
END
SELECT level_no, [name], total_price, child FROM @result
ORDER BY level_no
DROP TABLE RecursiveTable
CREATE TABLE NestedTable (lft INT NOT NULL, rgt INT NOT NULL, [Name]
[varchar] (20) COLLATE Hebrew_CI_AS NOT NULL ,
[Price] [int] NOT NULL)
INSERT INTO NestedTable (lft, rgt, [name], Price) VALUES(1,12, 'Dani',0)
INSERT INTO NestedTable (lft, rgt, [name], Price) VALUES(2,5, 'Jonn',100)
INSERT INTO NestedTable (lft, rgt, [name], Price) VALUES(3,4,'Toni',400)
INSERT INTO NestedTable (lft, rgt, [name], Price) VALUES(6,11,'Dorit',250)
INSERT INTO NestedTable (lft, rgt, [name], Price) VALUES(7,8,'Kimon',150)
INSERT INTO NestedTable (lft, rgt, [name], Price) VALUES(9,10,'Popo',200)
SELECT (SELECT COUNT(n1.lft) FROM NestedTable n1
WHERE n.lft BETWEEN n1.lft AND n1.rgt) ,
n.[name],
(SELECT SUM(n2.price) FROM NestedTable n2
WHERE n2.lft BETWEEN n.lft AND n.rgt)
FROM NestedTable n
ORDER BY n.lft
DROP TABLE NestedTable
--
Jacco Schalkwijk MCDBA, MCSD, MCSE
Database Administrator
Eurostop Ltd.
"Itzik" <it...@pisgasys.co.il> wrote in message
news:OQ4Zzzef...@TK2MSFTNGP11.phx.gbl...
Cristian Babu
============================================================================
create FUNCTION Price (@ID int)
RETURNS int
AS
BEGIN
declare @Price int, @ReturnValue int
set @Price = (select Price from RecursiveTable where Child = @ID)
set @ReturnValue = @Price
if exists(select * from RecursiveTable where Parent = @ID)
begin
declare @ChildID int
declare crs cursor local static forward_only for
select Child from RecursiveTable where Parent = @ID
open crs
fetch next from crs into @ChildID
while @@fetch_status = 0
begin
set @ReturnValue = @ReturnValue + dbo.Price(@ChildID)
fetch next from crs into @ChildID
end
close crs
deallocate crs
end
return(@ReturnValue)
END
go
select Parent, Child, Name, dbo.Price(Child) from recursivetable
order by Parent, Child
============================================================================
"Itzik" <it...@pisgasys.co.il> wrote in message
news:OQ4Zzzef...@TK2MSFTNGP11.phx.gbl...
Cristian Babu, CCNA
cristi...@softwaresolutions.ro
Alejandro Mesa
[Parent] [int] NOT NULL ,
[Child] [int] NOT NULL ,
[Price] [int] NOT NULL
) ON [PRIMARY]
GO
INSERT INTO RecursiveTable VALUES (1,2,'Jonn',100)
INSERT INTO RecursiveTable VALUES (1,3,'Dorit',250)
INSERT INTO RecursiveTable VALUES (3,4,'Kimon',150)
INSERT INTO RecursiveTable VALUES (2,5,'Toni',400)
INSERT INTO RecursiveTable VALUES (3,6,'Popo',200)
go
@child int
)
returns varchar(8000)
as
begin
declare @pk varchar(8000)
declare @parent int
declare @loop int
set @pk = ''
begin
set @parent = (select parent from dbo.recursiveTable where child = @loop)
begin
set @pk = cast(@parent as varchar) + '|' + @pk
end
end
end
go
as
select parent,
child,
[name],
price,
dbo.fnMyFunc(child) as tree_pk
from dbo.recursiveTable
go
a.[name],
sum(b.price) as sum_price
from vwMyview a
inner join
vwMyView b
on charindex(a.tree_pk, b.tree_pk, 1) = 1
group by a.tree_pk, len(a.tree_pk) - len(replace(a.tree_pk, '|', '')), a.[name]
order by a.tree_pk
go
drop function fnMyFunc
drop table recursiveTable
go
----------- -------------------- -----------
2 Jonn 500
3 Toni 400
2 Dorit 600
3 Kimon 150
3 Popo 200
Itzik
============================================================================
if exists (select * from dbo.sysobjects where id =
object_id(N'[dbo].[RecursiveTable]') and OBJECTPROPERTY(id, N'IsUserTable')
= 1)
drop table [dbo].[RecursiveTable]
GO
CREATE TABLE [dbo].[RecursiveTable] (
[Parent] [int] NOT NULL ,
[Child] [int] NOT NULL ,
[Name] [varchar] (20) COLLATE Hebrew_CI_AS NOT NULL ,
[Price] [int] NOT NULL
) ON [PRIMARY]
GO
INSERT INTO RecursiveTable VALUES (0,1,'Dani',0)
INSERT INTO RecursiveTable VALUES (1,2,'Jonn',100)
INSERT INTO RecursiveTable VALUES (1,3,'Dorit',250)
INSERT INTO RecursiveTable VALUES (3,4,'Kimon',150)
INSERT INTO RecursiveTable VALUES (2,5,'Toni',400)
INSERT INTO RecursiveTable VALUES (3,6,'Popo',200)
============================================================================
=======
I need solution to build Hierarchy tree and calculate data
this data will be receive :
1 Dani 1100
2 Jonn 500
3 Toni 400
2 Dorit 600
3 Kimon 150
3 Popo 200
Thanks, Itzik
John Bell
To extend on Jacco's reply on 18-Sept-2003
Please don't post the same question three times. If are still having
problems with this solution add the the existing thread.
See
John
"Itzik" <it...@pisgasys.co.il> wrote in message
news:OIK$caAgDH...@TK2MSFTNGP11.phx.gbl...
Joe Celko
called an adjacency list model and it looks like this:
CREATE TABLE OrgChart
(emp CHAR(10) NOT NULL PRIMARY KEY,
boss CHAR(10) DEFAULT NULL REFERENCES OrgChart(emp),
salary DECIMAL(6,2) NOT NULL DEFAULT 100.00);
OrgChart
emp boss salary
===========================
'Albert' 'NULL' 1000.00
'Bert' 'Albert' 900.00
'Chuck' 'Albert' 900.00
'Donna' 'Chuck' 800.00
'Eddie' 'Chuck' 700.00
'Fred' 'Chuck' 600.00
Another way of representing trees is to show them as nested sets.
Since SQL is a set oriented language, this is a better model than the
usual adjacency list approach you see in most text books. Let us define
a simple OrgChart table like this.
CREATE TABLE OrgChart
(emp CHAR(10) NOT NULL PRIMARY KEY,
lft INTEGER NOT NULL UNIQUE CHECK (lft > 0),
rgt INTEGER NOT NULL UNIQUE CHECK (rgt > 1),
CONSTRAINT order_okay CHECK (lft < rgt) );
OrgChart
emp lft rgt
======================
'Albert' 1 12
'Bert' 2 3
'Chuck' 4 11
'Donna' 5 6
'Eddie' 7 8
'Fred' 9 10
The organizational chart would look like this as a directed graph:
Albert (1, 12)
/ \
/ \
Bert (2, 3) Chuck (4, 11)
/ | \
/ | \
/ | \
/ | \
Donna (5, 6) Eddie (7, 8) Fred (9, 10)
The adjacency list table is denormalized in several ways. We are
modeling both the Personnel and the organizational chart in one table.
But for the sake of saving space, pretend that the names are job titles
and that we have another table which describes the Personnel that hold
those positions.
Another problem with the adjacency list model is that the boss and
employee columns are the same kind of thing (i.e. names of personnel),
and therefore should be shown in only one column in a normalized table.
To prove that this is not normalized, assume that "Chuck" changes his
name to "Charles"; you have to change his name in both columns and
several places. The defining characteristic of a normalized table is
that you have one fact, one place, one time.
The final problem is that the adjacency list model does not model
subordination. Authority flows downhill in a hierarchy, but If I fire
Chuck, I disconnect all of his subordinates from Albert. There are
situations (i.e. water pipes) where this is true, but that is not the
expected situation in this case.
To show a tree as nested sets, replace the nodes with ovals, and then
nest subordinate ovals inside each other. The root will be the largest
oval and will contain every other node. The leaf nodes will be the
innermost ovals with nothing else inside them and the nesting will show
the hierarchical relationship. The (lft, rgt) columns (I cannot use the
reserved words LEFT and RIGHT in SQL) are what show the nesting. This is
like XML, HTML or parentheses.
At this point, the boss column is both redundant and denormalized, so it
can be dropped. Also, note that the tree structure can be kept in one
table and all the information about a node can be put in a second table
and they can be joined on employee number for queries.
To convert the graph into a nested sets model think of a little worm
crawling along the tree. The worm starts at the top, the root, makes a
complete trip around the tree. When he comes to a node, he puts a number
in the cell on the side that he is visiting and increments his counter.
Each node will get two numbers, one of the right side and one for the
left. Computer Science majors will recognize this as a modified preorder
tree traversal algorithm. Finally, drop the unneeded OrgChart.boss
column which used to represent the edges of a graph.
This has some predictable results that we can use for building queries.
The root is always (left = 1, right = 2 * (SELECT COUNT(*) FROM
TreeTable)); leaf nodes always have (left + 1 = right); subtrees are
defined by the BETWEEN predicate; etc. Here are two common queries which
can be used to build others:
1. An employee and all their Supervisors, no matter how deep the tree.
SELECT O2.*
FROM OrgChart AS O1, OrgChart AS O2
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
AND O1.emp = :myemployee;
2. The employee and all subordinates. There is a nice symmetry here.
SELECT O1.*
FROM OrgChart AS O1, OrgChart AS O2
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
AND O2.emp = :myemployee;
3. Add a GROUP BY and aggregate functions to these basic queries and you
have hierarchical reports. For example, the total salaries which each
employee controls:
SELECT O2.emp, SUM(S1.salary)
FROM OrgChart AS O1, OrgChart AS O2,
Salaries AS S1
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
AND O1.emp = S1.emp
GROUP BY O2.emp;
4. To find the level of each emp, so you can print the tree as an
indented listing. Technically, you should declare a cursor to go with
the ORDER BY clause.
SELECT COUNT(O2.emp) AS indentation, O1.emp
FROM OrgChart AS O1, OrgChart AS O2
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
GROUP BY O1.lft, O1.emp
ORDER BY O1.lft;
5. The nested set model has an implied ordering of siblings which the
adjacency list model does not. To insert a new node, G1, under part G.
We can insert one node at a time like this:
BEGIN ATOMIC
DECLARE rightmost_spread INTEGER;
SET rightmost_spread
= (SELECT rgt
FROM Frammis
WHERE part = 'G');
UPDATE Frammis
SET lft = CASE WHEN lft > rightmost_spread
THEN lft + 2
ELSE lft END,
rgt = CASE WHEN rgt >= rightmost_spread
THEN rgt + 2
ELSE rgt END
WHERE rgt >= rightmost_spread;
INSERT INTO Frammis (part, lft, rgt)
VALUES ('G1', rightmost_spread, (rightmost_spread + 1));
COMMIT WORK;
END;
The idea is to spread the (lft, rgt) numbers after the youngest child of
the parent, G in this case, over by two to make room for the new
addition, G1. This procedure will add the new node to the rightmost
child position, which helps to preserve the idea of an age order among
the siblings.
6. To convert a nested sets model into an adjacency list model:
SELECT B.emp AS boss, E.emp
FROM OrgChart AS E
LEFT OUTER JOIN
OrgChart AS B
ON B.lft
= (SELECT MAX(lft)
FROM OrgChart AS S
WHERE E.lft > S.lft
AND E.lft < S.rgt);
7. To convert an adjacency list to a nested set model, use a push down
stack. Here is version with a stack in SQL/PSM.
-- Tree holds the adjacency model
CREATE TABLE Tree
(node CHAR(10) NOT NULL,
parent CHAR(10));
-- Stack starts empty, will holds the nested set model
CREATE TABLE Stack
(stack_top INTEGER NOT NULL,
node CHAR(10) NOT NULL,
lft INTEGER,
rgt INTEGER);
CREATE PROCEDURE TreeTraversal ()
LANGUAGE SQL
DETERMINISTIC
BEGIN ATOMIC
DECLARE counter INTEGER;
DECLARE max_counter INTEGER;
DECLARE current_top INTEGER;
SET counter = 2;
SET max_counter = 2 * (SELECT COUNT(*) FROM Tree);
SET current_top = 1;
--clear the stack
DELETE FROM Stack;
-- push the root
INSERT INTO Stack
SELECT 1, node, 1, max_counter
FROM Tree
WHERE parent IS NULL;
-- delete rows from tree as they are used
DELETE FROM Tree WHERE parent IS NULL;
WHILE counter <= max_counter- 1
DO IF EXISTS (SELECT *
FROM Stack AS S1, Tree AS T1
WHERE S1.node = T1.parent
AND S1.stack_top = current_top)
THEN BEGIN -- push when top has subordinates and set lft value
INSERT INTO Stack
SELECT (current_top + 1), MIN(T1.node), counter, NULL
FROM Stack AS S1, Tree AS T1
WHERE S1.node = T1.parent
AND S1.stack_top = current_top;
-- delete rows from tree as they are used
DELETE FROM Tree
WHERE node = (SELECT node
FROM Stack
WHERE stack_top = current_top + 1);
-- housekeeping of stack pointers and counter
SET counter = counter + 1;
SET current_top = current_top + 1;
END;
ELSE
BEGIN -- pop the stack and set rgt value
UPDATE Stack
SET rgt = counter,
stack_top = -stack_top -- pops the stack
WHERE stack_top = current_top;
SET counter = counter + 1;
SET current_top = current_top - 1;
END;
END IF;
-- the top column is not needed in the final answer
END WHILE;
-- SELECT node, lft, rgt FROM Stack;
END;
I have a book on TREES & HIERARCHIES IN SQL coming out in April 2004.
--CELKO--
===========================
Please post DDL, so that people do not have to guess what the keys,
constraints, Declarative Referential Integrity, datatypes, etc. in your
schema are.
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