It is on Amazon.com. here is the whole list:

Joe Celko's Data and Databases : Concepts in Practice 1-55860-432-4

Joe Celko's SQL Programming Style 0-12-088797-5

Joe Celko's SQL Puzzles and Answers 5 1-55860-453-7

Joe Celko's Trees and Hierarchies in SQL for Smarties $34.95

1-55860-920-2

Joe Celko's SQL for Smarties:Advanced SQL Programming Second Edition

1-55860-576-2

Joe Celko's SQL for Smarties (3rd Edition): Advanced SQL Programming

Joe Celko 0-12-369379-9

but I was unable to find a way to get a reference to an identity field [sic] in the same record[sic] <<

Again, if you keep using the wrong terms, you will never see how write

code in a declarative relational language.

Nexr, you should never use IDENTIT. It is not relational and **by

definition** cannot be a key.

There are many ways to represent a tree or hierarchy in SQL. This 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.

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 their 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;

END WHILE;

-- SELECT node, lft, rgt FROM Stack;

-- the top column is not needed in the final answer

-- move stack contents to new tree table

END;