Author 
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AskSQLTeam
Ask SQLTeam Question
USA
0 Posts 
Posted  03/21/2002 : 00:29:01

Scott writes "I have been doing a lot of recursive selects lately, but one that seems to elude me, is the most effective way to do a recursive Join on a table with the following design:
ID  ParentID  Section
Which would contatin 1  null  Programming 2  1  VBScript 3  2  Tutorials etc.
So that the stored procedure returns data like this:
ID  Recursive Section Listing
Such as:
1  Programming 2  Programming > VBScript 3  Programming > VBScript > Tutorials 4  Programming > Java
etc.
This would then allow me to do some nice bread crumbs, and to put in a drop down menu for users to select.
Thanks! Scott." 

AndrewMurphy
Flowing Fount of Yak Knowledge
Ireland
2916 Posts 
Posted  03/21/2002 : 04:34:27

search here for items on "heirarchy" or "treestructure"...i think a solution to your problem might already exist.



rrb
SQLTeam Poet Laureate
Australia
1479 Posts 

jcelko
Esteemed SQL Purist
USA
547 Posts 
Posted  03/22/2002 : 11:45:40

The usual example of a tree structure in SQL books is called an adjacency list model and it looks like this:
CREATE TABLE Personnel (emp CHAR(10) NOT NULL PRIMARY KEY, boss CHAR(10) DEFAULT NULL REFERENCES Personnel(emp), salary DECIMAL(6,2) NOT NULL DEFAULT 100.00);
Personnel 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 Personnel table like this, ignoring the left (lft) and right (rgt) columns for now. This problem is always given with a column for the employee and one for his boss in the textbooks. This table without the lft and rgt columns is called the adjacency list model, after the graph theory technique of the same name; the pairs of nodes are adjacent to each other.
CREATE TABLE Personnel (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) );
Personnel 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 first 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, 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 rgt and lft columns (I cannot use the reserved words LEFT and RIGHT in SQL) are what shows the nesting.
If that mental model does not work, then imagine a little worm crawling anticlockwise along the tree. Every time he gets to the left or right side of a node, he numbers it. The worm stops when he gets all the way around the tree and back to the top.
This is a natural way to model a parts explosion, since a final assembly is made of physically nested assemblies that final break down into separate parts.
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 Personnel.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 P2.* FROM Personnel AS P1, Personnel AS P2 WHERE P1.lft BETWEEN P2.lft AND P2.rgt AND P1.emp = :myemployee;
2. The employee and all subordinates. There is a nice symmetry here.
SELECT P1.* FROM Personnel AS P1, Personnel AS P2 WHERE P1.lft BETWEEN P2.lft AND P2.rgt AND P2.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 P2.emp, SUM(S1.salary) FROM Personnel AS P1, Personnel AS P2, Salaries AS S1 WHERE P1.lft BETWEEN P2.lft AND P2.rgt AND P1.emp = S1.emp GROUP BY P2.emp;
4. To find the level of each node, so you can print the tree as an indented listing.
DECLARE Out_Tree CURSOR FOR SELECT P1.lft, COUNT(P2.emp) AS indentation, P1.emp FROM Personnel AS P1, Personnel AS P2 WHERE P1.lft BETWEEN P2.lft AND P2.rgt GROUP BY P1.emp ORDER BY P1.lft;
5. The nested set model has an implied ordering of siblings which the adjacency list model does not. To insert a new node as the rightmost sibling.
BEGIN DECLARE right_most_sibling INTEGER;
SET right_most_sibling = (SELECT rgt FROM Personnel WHERE emp = :your_boss);
UPDATE Personnel SET lft = CASE WHEN lft > right_most_sibling THEN lft + 2 ELSE lft END, rgt = CASE WHEN rgt >= right_most_sibling THEN rgt + 2 ELSE rgt END WHERE rgt >= right_most_sibling;
INSERT INTO Personnel (emp, lft, rgt) VALUES ('New Guy', right_most_sibling, (right_most_sibling + 1)) END;
6. To convert a nested sets model into an adjacency list model:
SELECT B.emp AS boss, P.emp FROM Personnel AS P LEFT OUTER JOIN Personnel AS B ON B.lft = (SELECT MAX(lft) FROM Personnel AS S WHERE P.lft > S.lft AND P.lft < S.rgt);
For details, see the chapter in my book JOE CELKO'S SQL FOR SMARTIES (MorganKaufmann, 1999, second edition)
http://searchdatabase.techtarget.com/tip/1,289483,sid13_gci537290,00.html
http://searchdatabase.techtarget.com/tip/1,289483,sid13_gci801943,00.html
CELKO Joe Celko, SQL Guru



AndrewMurphy
Flowing Fount of Yak Knowledge
Ireland
2916 Posts 
Posted  12/12/2002 : 08:09:56

just working through Joe's contribution at the bottom......
item 4. "To find the level of each node, so you can print the tree as an indented listing. "
SELECT P1.lft, COUNT(P2.emp) AS indentation, P1.emp FROM Personnel AS P1, Personnel AS P2 WHERE P1.lft BETWEEN P2.lft AND P2.rgt GROUP BY P1.emp ORDER BY P1.lft;
as written above doesn't work....it SHOULD READ...
SELECT P1.lft, COUNT(P2.emp) AS indentation, P1.emp FROM Personnel AS P1, Personnel AS P2 WHERE P1.lft BETWEEN P2.lft AND P2.rgt GROUP BY P1.emp, p1.lft ORDER BY P1.lft;
Hope the above helps anybody dipping their toes into this subject like I am currently.....now I must move on to the next level of difficulty.



neeraj_sharma
Starting Member
India
15 Posts 
Posted  06/12/2013 : 06:41:27

Thanks for this thread i was looking for it :)
Neeraj Prasad
Tutorial Sql Server 


visakh16
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