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Tip of the Day
Language: SQL
Expertise: Intermediate
Oct 5, 2004

Hierarchical Tree Structures in Database Tables

This recursive function retrieves a hierarchical tree structure from a table with just four columns that describe each node in the tree: Level (the depth of the node in the tree), ID, Name, and ParentID. Given an ID and a Level, the function opens a cursor and retrieves the entire branch of the tree for the specified node at that level.

First, create the table:


Create Table Tree(ID int,Name Varchar(10),ParentID int)
Go
Next, create the function:

Create Function getTree(@ID Int,@Level int=1)
Returns @Tree Table(Level int,ID int,Name Varchar(10),ParentID int)
As
Begin
	Declare @tID int,@tName Varchar(10),@tParentID int
	Declare TreeCur Cursor
	For	Select * From Tree Where ParentID=@ID
	Open TreeCur
	FETCH NEXT FROM TreeCur into @tID,@tName,@tParentID
	Insert Into @Tree Select @Level,* From Tree Where ID=@ID
	Set @Level=@Level+1
	WHILE (@@fetch_status = 0)
	Begin
		Insert Into @Tree Select * From dbo.getTree(@tID,@Level)
		FETCH NEXT FROM TreeCur into @tID,@tName,@tParentID
	End
	Close TreeCur
	Deallocate TreeCur
	return
End
Lastly, fill the table:

INSERT INTO Tree  VALUES(1,'Root',0)
Go
INSERT INTO Tree  VALUES(2,'Node 1',1)
Go
INSERT INTO Tree  VALUES(3,'Node 2',1)
Go
INSERT INTO Tree  VALUES(4,'Node 1.1',2)
Go
INSERT INTO Tree  VALUES(5,'Node 1.2',2)
Go
INSERT INTO Tree  VALUES(6,'Node 1.1.1',4)
Go
INSERT INTO Tree  VALUES(7,'Node 2.1',3)
Go
Build the tree using Function
To get the tree built from the root, use this:

select * from getTree(1,DEFAULT)

Level       ID          Name       ParentID    
----------- ----------- ---------- ----------- 
1           1           Root       0
2           2           Node 1     1
3           4           Node 1.1   2
4           6           Node 1.1.1 4
3           5           Node 1.2   2
2           3           Node 2     1
3           7           Node 2.1   3

(7 row(s) affected)
To retrieve the tree built from Node 1, use:

select * from getTree(2,DEFAULT)

Level       ID          Name       ParentID    
----------- ----------- ---------- ----------- 
1           2           Node 1     1
2           4           Node 1.1   2
3           6           Node 1.1.1 4
2           5           Node 1.2   2

(4 row(s) affected)
To retrieve the tree built from Node 2, use:

select * from getTree(3,DEFAULT)

Level       ID          Name       ParentID    
----------- ----------- ---------- ----------- 
1           3           Node 2     1
2           7           Node 2.1   3

(2 row(s) affected)
Rohit K. Gupta
 
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