>> and if I delete a row that has parent=0 I want to delete those who

has parent=id for the one deleted <<

I think you are still using the wrong representation for a tree in SQL

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

anti-clockwise 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:

0. Your problem was to delete an entire subtree:

DELETE FROM Personnel

WHERE Personnel.lft

BETWEEN (SELECT P1.lft

FROM Personnel AS P1

WHERE P1.emp = :myemployee)

AND (SELECT P2.rgt

FROM Personnel AS P2

WHERE P2.emp = :myemployee);

This one statement will work for ANY depth of subtree.

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 P2.*

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.

SELECT 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 an adjacency list model into a nested set model, use a

push down stack algorithm. Assume that we have these tables:

-- Tree holds the adjacency model

CREATE TABLE Tree

(emp CHAR(10) NOT NULL,

boss CHAR(10));

INSERT INTO Tree

SELECT emp, boss FROM Personnel;

-- Stack starts empty, will holds the nested set model

CREATE TABLE Stack

(stack_top INTEGER NOT NULL,

emp CHAR(10) NOT NULL,

lft INTEGER,

rgt INTEGER);

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;

INSERT INTO Stack

SELECT 1, emp, 1, NULL

FROM Tree

WHERE boss IS NULL;

DELETE FROM Tree

WHERE boss IS NULL;

WHILE counter <= (max_counter - 2)

LOOP IF EXISTS (SELECT *

FROM Stack AS S1, Tree AS T1

WHERE S1.emp = T1.boss

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.emp), counter, NULL

FROM Stack AS S1, Tree AS T1

WHERE S1.emp = T1.boss

AND S1.stack_top = current_top;

DELETE FROM Tree

WHERE emp = (SELECT emp

FROM Stack

WHERE stack_top = current_top + 1);

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 IF;

END LOOP;

END;

This approach will be two to three orders of magnitude faster than the

adjacency list model for subtree and aggregate operations.

For details, see the chapter in my book JOE CELKO'S SQL FOR SMARTIES

(Morgan-Kaufmann, 1999, second edition)

--CELKO--

Joe Celko, SQL and Database Consultant

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