I am working on a data model that uses a nested set in MySQL. I'm semi-
aware of the relational and other shortcomings of MySQL; please
forgive me if discussion of the platform is verboten here. I have a
problem that I believe is a general RDBMS problem; if it is instead a
MySQL problem, please let me know and give me an idea of the specific
solution as it might be implemented in another RDMS

We have a table Clients that's a list of clients. We have a table of
Units which is retail units in a physical location. The simple model
is to give each unit a client_id, but there is a hierarchical grouping
system that we need to do reports by, so a simple one-to-many won't do
it.

For any client, a unit might belong under various organizational
levels. A unit might be a corporate unit, so it's simply under
corporate. A unit might belong to a franchise, so the org chart would
go something like Corporate - Franchise A - Unit 12 . A unit might be
in the New York district of the Eastern Region, so the org chart there
would look like Corporate - Eastern Region - New York District - Unit
36. In this case, the Eastern Region wouldn't have any Units that
belong solely under it; it's simply a 'virtual' grouping level.

Any client might have any kind of organizational chart, with whatever
naming schema. I'm going to assume for now that they're all strictly
hierarchical -- no overlapping. And, there may be any number of
virtual grouping levels.

So, now I will make a ClientOrgChart, which will be used to join
Clients to Units. I've been looking at both adjacency list and the
preorder tree traversal algorithm from this web page:
http://dev.mysql.com/tech-resources/...ical-data.html .

But, now I'm not sure the best way to relate the ClientOrgChart to
clients. In any of the ways I imagine it, I can see ambiguity arising,
and I don't see any way to prevent it using the DDL. I'll use the
adjacency list in the examples since it's easier to type out.

Suppose I had this adjacency list:

id client_id parent_id level
1 1 NULL Corporate
2 1 1 Franchise 1
4 1 1 Eastern Region
5 1 4 New York District

All child nodes *should* have the same client_id as the ultimate root,
but I can't think of any way to ensure this in the DDL. I'll just have
to "be careful" in code.

A benefit of this is that I can easily join a unit to the client
table, by looking that the client_id of the ClientOrgChart record that
it's related to, but again, there's a ( hopefully small ) risk of it
being wrong.

Or, I could make the client_id allow a NULL value

id client_id parent_id level
1 1 NULL Corporate
2 NULL 1 Franchise 1
4 NULL 1 Eastern Region
5 NULL 4 New York District

So have I fixed anything? I can't enforce a rule like 'Either
client_id or parent_id must be null'. Figuring out the client that a
unit ultimately belongs to would be a slightly more complex query,
also.

Another way to do it would be to make a one-to-one table ( never
thought I'd use that ) that joins Clients to ClientOrgChart, but
there's no way to prevent the insertion of a non-root node value in
this table... same problem.

The usual solutions that my mind jumps to all seem not really to solve
the underlying problem -- possible conflicting client_id values. Is
this problem intractable? Is the best solution a compromise, not a
perfect one?

lawpoop wrote:
....
Can't comment on product techniques but if this were a 'table', I'd
wonder how a client can be its own parent. I would have thought some
constraint would prevent that.

On Apr 13, 9:27*am, paul c <toledobythe... (AT) oohay (DOT) ac> wrote:
Are you asking why all of these example rows have a value for the
client_id? Are you saying it should look like this? :

id client_id parent_id level
1 1 NULL Corporate
2 NULL 1 Franchise 1
4 NULL 1 Eastern Region
5 NULL 4 New York District

MySQL check constraints are somewhat limited; what would you suggest?

lawpoop wrote:
No, assuming you plan to recurse a tree, how will you end the recursion?

Can't help with MySQL.

On Apr 13, 10:29*am, paul c <toledobythe... (AT) oohay (DOT) ac> wrote:

Can you show me what you think is going to happen? I've done this
before, and it works fine. Recursion ends when you get either to the
root node, or when you get to the last terminal node, depending on
which direction you are going, up or down.

The column parent_id is related to id, not client_id. It's a self-
join. See 'The Adjacency List model' in this article:
http://dev.mysql.com/tech-resources/...ical-data.html

Now that I'm working with this, it looks like each record needs its
own client_id value. Otherwise, I don't have a way of specifying which
tree to update when I'm inserting a new node.

lawpoop wrote:
....
Oh, sorry, I missed that.

[Snipped]


Where is the surprise ?
After all, it was only until recently that MySQL would reject
'31/13/2009' as valid date when using a date/time data type.

First, buy a ccopy of TREES & HIERARCHIES IN SQL. It will five you a
lot more infomatuin than a Newsgroup. Then it would help if you
would post real DDL instead a narrative.

We have a table Clients that's a list of clients. We have a table of
Units which is retail units in a physical location. The simple model
is to give each unit a client_id, but there is a hierarchical grouping
system that we need to do reports by, so a simple one-to-many won't do
it.

CREATE TABLE Clients
(client_id CHAR(9) NOT NULL PRIMARY KEY, -- duns number?
client_name VARCHAR(15) NOT NULL,
... );


Here is a skeleton for the Nested Sets model

CREATE TABLE ClientOrgCharts
(client_id CHAR(9) NOT NULL UNIQUE
REFERENCES Client(client_id),
lft INTEGER NOT NULL UNIQUE CHECK (lft > 0),
rgt INTEGER NOT NULL UNIQUE CHECK (rgt > 1),
CONSTRAINT order_okay CHECK (lft < rgt));

Look at the Nested Sets skeleton. No cycles since each client_id
appears once. In a full SQL you can also add more constriants to
assure a proper tree strcuture. Or use a VIEW and a WITH CHECK
OPTION.

The adjacency list reqiures cursors to do data integrity. Here is my
"cut & paste" on tress:

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_name CHAR(10) NOT NULL PRIMARY KEY,
boss_emp_name CHAR(10) REFERENCES OrgChart(emp_name),
salary_amt DECIMAL(6,2) DEFAULT 100.00 NOT NULL,
<< horrible cycle constaints >>);

OrgChart
emp_name boss_emp_name salary_amt
==============================
'Albert' NULL 1000.00
'Bert' 'Albert' 900.00
'Chuck' 'Albert' 900.00
'Donna' 'Chuck' 800.00
'Eddie' 'Chuck' 700.00
'Fred' 'Chuck' 600.00

This approach will wind up with really ugly code -- CTEs hiding
recursive procedures, horrible cycle prevention code, etc. The root
of your problem is not knowing that rows are not records, that SQL
uses sets and trying to fake pointer chains with some vague, magical
non-relational "id".

This matches the way we did it in old file systems with pointer
chains. Non-RDBMS programmers are comfortable with it because it
looks familiar -- it looks like records and not rows.

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_name 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_name lft rgt
======================
'Albert' 1 12
'Bert' 2 3
'Chuck' 4 11
'Donna' 5 6
'Eddie' 7 8
'Fred' 9 10

The (lft, rgt) pairs are like tags in a mark-up language, or parens in
algebra, BEGIN-END blocks in Algol-family programming languages, etc.
-- they bracket a sub-set. This is a set-oriented approach to trees
in a set-oriented language.

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_emp_name and employee columns are the same kind of thing (i.e.
identifiers 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_emp_name 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_emp_name 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_name = :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_name = :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_name, SUM(S1.salary_amt)
FROM OrgChart AS O1, OrgChart AS O2,
Salaries AS S1
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
AND O1.emp_name = S1.emp_name
GROUP BY O2.emp_name;

4. To find the level of each emp_name, so you can print the tree as an
indented listing.

SELECT T1.node,
SUM(CASE WHEN T2.lft <= T1.lft THEN 1 ELSE 0 END
+ CASE WHEN T2.rgt < T1.lft THEN -1 ELSE 0 END) AS lvl
FROM Tree AS T1, Tree AS T2
WHERE T2.lft <= T1.lft
GROUP BY T1.node;

An untested version of this using OLAP functions might be better able
to use the ordering.

SELECT T1.node,
SUM(CASE WHEN T2.lft <= T1.lft THEN 1 ELSE 0 END
+ CASE WHEN T2.rgt < T1.lft THEN -1 ELSE 0 END)
OVER (ORDER BY T1.lft
RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS
lvl
FROM Tree AS T1, Tree AS T2
WHERE T2.lft <= T1.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_name AS boss_emp_name, E.emp_name
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 find the immediate parent of a node:

SELECT MAX(P2.lft), MIN(P2.rgt)
FROM Personnel AS P!, Personnel AS P2
WHERE P1.lft BETWEEN P2.lft AND P2.rgt
AND P1.emp_name = :my_emp_name;

8. 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 O1.
WHERE S1.node = O1.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(O1.node), counter, NULL
FROM Stack AS S1, Tree AS O1.
WHERE S1.node = O1.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;

I have a book on TREES & HIERARCHIES IN SQL which you can get at
Amazon.com right now.

For a good article on using CTEs and recursion, see:
http://www.sqlservercentral.com/colu...server2005.asp

On Apr 14, 6:18*am, --CELKO-- <jcelko... (AT) earthlink (DOT) net> wrote:
Thanks, Celko, your articles have been exceedingly helpful for me in
understanding this problem. I'll pick up a copy of your book!

The actual Clients table DDL that we're using has a bit more stuff in
it, but for this problem, all that's relevant is basically what you
posted:

CREATE TABLE IF NOT EXISTS `Clients` (
`id` int(10) unsigned NOT NULL auto_increment,
`client_name` varchar(40) NOT NULL default '',
...
PRIMARY KEY (`id`),
) ENGINE=InnoDB DEFAULT CHARSET=utf8 ;

The Locations table looks similar, for this purpose:

CREATE TABLE IF NOT EXISTS `Locations` (
`id` int(10) unsigned NOT NULL auto_increment,
`client_id` int(10) unsigned NOT NULL,
`location_name` varchar(30) character set utf8 NOT NULL default '',
`unit_number` varchar(30) character set utf8 NOT NULL,
...
PRIMARY KEY (`id`),
) ENGINE=InnoDB DEFAULT CHARSET=utf8 ;

ALTER TABLE `Locations`
ADD CONSTRAINT `Locations_ibfk_1` FOREIGN KEY (`client_id`)
REFERENCES `Clients` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
)

It used to have client_id as a foreign key, but it will change to
have org_chart_id as a foreign key, to describe its place in the
hierarchy.

ALTER TABLE `Locations`
ADD CONSTRAINT `Locations_ibfk_1` FOREIGN KEY (`org_chart_id`)
REFERENCES `ClientOrgChart` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
)


Now for the ClientOrgChart. I decided to go with a modified preorder
tree traversal algorithm:

CREATE TABLE IF NOT EXISTS `ClientOrgChart` (
`id` int(10) unsigned NOT NULL auto_increment,
`client_id` int(10) unsigned NOT NULL,
`level` varchar(50) collate utf8_unicode_ci NOT NULL COMMENT 'the
name of the organizational level', e.g. ''Eastern Region''',
`lft` int(10) unsigned NOT NULL,
`rgt` int(10) unsigned NOT NULL,
PRIMARY KEY (`id`),
UNIQUE KEY `client_id` (`client_id`,`lft`,`rgt`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci
COMMENT='This is a modified preorder tree traversal algorithm.' ;


ALTER TABLE `ClientOrgChart`
ADD CONSTRAINT `ClientOrgChart_ibfk_1` FOREIGN KEY (`client_id`)
REFERENCES `Clients` (`id`) ON DELETE CASCADE ON UPDATE CASCADE;

What was bugging me was the client_id -- it seemed to me that I really
should only need to know it in one place, at the root node, like a
parent/child relationship. But actually, in this scheme, it's a way of
differentiating one tree from another, since the nodes of all trees
are mixed together in a single table.

Since I'm using MySQL, I'm limited in the SQL that I have available.