How to create a newsgroup-like db
arthur...@my-deja.com
I would like to create a table in a database that contains messages
that are structures like a newsgroup; with replies and threads. Can
anyone give a good example on what the structure of the table should
look like?
When I retrieve the list of messages from the table I want to order
them exactly the right way, such as this:
a. Message One
b. Re: Message One
c. Re: Re: Message One
d. Re: Message One [2]
e. Re: Re: Message One [2]
I guess in the database I have to give a thread an ID number and then
at least sort the list by thread. But still, how do I fit in replies
that are placed later inbetween other replies? Do I have to use
an 'Order' field in my database and update all those fields in the
entire thread when a reply is placed?
I hope someone here can give me an idea on how to do this.
Greetings,
Arthur
Sent via Deja.com http://www.deja.com/
Before you buy.
joe_...@my-deja.com
>> I would like to create a table in a database that contains messages
that are structures like a newsgroup; with replies and threads. <<
Oh no! You asked a quesiton about a tree structure! That means I have
to post my famous tree file <g>!!
===================
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 is a bad example, since it combines Personnel and Organizational
Chart data into one table. I did it this way to keep the post small;
imagine that the names are job titles if you want to be picky.
CREATE TABLE Personnel
(emp CHAR(10) PRIMARY KEY,
boss CHAR(10), -- this column is unneeded & denormalizes the
table
salary DECIMAL(6,2) NOT NULL,
lft INTEGER NOT NULL,
rgt INTEGER NOT NULL);
Personnel
emp boss salary lft rgt
===================================
Albert NULL 1000.00 1 12
Bert Albert 900.00 2 3
Chuck Albert 900.00 4 11
Donna Chuck 800.00 5 6
Eddie Chuck 700.00 7 8
Fred Chuck 600.00 9 10
which 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)
This (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. The 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.
What we should have is one table for the organizational chart and one
for the personnel, so that you can separate people from their
positions. Ignore that and let me use the names as if they are
employee identifiers.
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:
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(P1.salary)
FROM Personnel AS P1, Personnel AS P2
WHERE P1.lft BETWEEN P2.lft AND P2.rgt
GROUP BY P2.emp;
4. The level of each node in the heirarchy is
SELECT COUNT(P1.emp) AS level, P2.emp
FROM Personnel AS P1, Personnel AS P2
WHERE P1.lft BETWEEN P2.lft AND P2.rgt
GROUP BY P2.emp
ORDER BY p2.lft;
This will print out the indented listing of the tree structure.
Nested set models will be two to three orders of magnitude faster than
the adjacency list model.
For details, see chapter 29 in my book JOE CELKO'S SQL FOR SMARTIES
second edition (Morgan-Kaufmann, 1999).
--CELKO--
arthur...@my-deja.com
Thank you for the great information.
One more question: How do I assign the right LFT and RGT values to each
record? What I have, for instance, is a table where each record has an
ID number and a field that says what the ID number is of the record
that is it's parent.
I can draw the tree on paper and fill in all the LFT's and RGT's
myself. But when someone would add a record somewhere I want to go
through the tree again and re-number the LFT and RGT fields.
Greetings,
Arthur
In article <8c2s9m$a7j$1...@nnrp1.deja.com>,
Joe Celko
>> One more question: How do I assign the right LFT and RGT values to
each record? <<
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;
INSERT INTO Stack
SELECT 1, emp, 1, NULL
FROM Tree
WHERE boss IS NULL;
DELETE FROM Tree
WHERE boss IS NULL;
SET counter = 2;
SET max_counter = 2 * (SELECT COUNT(*) FROM Tree) - 2;
SET current_top = 1;
WHILE counter <= max_counter
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
WHERE stack_top = current_top
SET counter = counter + 1;
SET current_top = current_top - 1;
END IF;
END LOOP;
END;
--CELKO--
Joe Celko, SQL and Database Consultant