Quote:>> I am looking for implementing a database which represents a

directory tree; <<

The usual example of a tree structure in SQL books is 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, 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

emps are adjacent to each other.

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 first table is denormalized in several ways. We are modeling both

the OrgChart 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 OrgChart 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 OrgChart),

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 emps with ovals, then nest

subordinate ovals inside each other. The root will be the largest oval

and will contain every other emp. The leaf emps 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 anti-clockwise along the tree. Every time he gets to the left

or right side of a emp, 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 emp 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 emp, he puts a

number in the cell on the side that he is visiting and increments his

counter. Each emp 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 emps 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.

DECLARE Out_Tree CURSOR FOR

SELECT O1.lft, 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.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 emp as the rightmost

sibling.

UPDATE OrgChart

SET lft = lft + 2,

rgt = rgt + 2

WHERE rgt >= (SELECT rgt -- right_most_sibling

FROM OrgChart

WHERE emp = :your_boss);

INSERT INTO OrgChart (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 OrgChart AS P

LEFT OUTER JOIN

OrgChart AS B

ON B.lft

= (SELECT MAX(lft)

FROM OrgChart 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

(Morgan-Kaufmann, 1999, second edition)

http://searchdatabase.techtarget.com/tip/1,289483,sid13_gci537290,00....

http://searchdatabase.techtarget.com/tip/1,289483,sid13_gci801943,00....