Hi!

There's a database with following tables:

1) transition closure - helper table that contains hierarchy description
with columns: node_id,node_parent_id,node_level.

2) table nodes - a table of nodes, directories or whatever - simply
members of hierarchy

3) table people - table containing users descriptions

the problem is about connecting users with hierarchy in a way, that
every user can have its own root directory.

imagine it with file system.
we have a directory, eg. "root". within this directory we have
a number of subdirectories that contain other, more and more nested,
directories ...

now, we have another direcory, "users".

In order to have user have his own root dir, we simply use aliases.
for example "users/john" has two aliases "apples" and "buckets" that
link to "root/apples" and "root/things/buckets". This is simple with
file system, unfortuanetly i need to have an sql solution.

I tried to add another table, ("people_root_contents") with column
node_id,people_id that contained users root dir contents, but as you can
easily imagine, the query for searching the whole user-dependant
hierarchy was huge and slow.


Any ideas?

Regards
H

Hitch wrote:
How are you "searching the whole user-dependant hierarchy"?

Are you using a recursive query?

--
Lee Fesperman, FirstSQL, Inc. (http://www.firstsql.com)
================================================== ============
* The Ultimate DBMS is here!
* FirstSQL/J Object/Relational DBMS (http://www.firstsql.com)

Lee Fesperman <firstsql (AT) ix (DOT) netcom.com> wrote:

No, i use joins. (pgSQL).

Regards
H

Hitch wrote:

Add a column "link_target" to your node table. If a node specifies a directory,
its NULL, if it is a alias/symbolic link, then enter the link end node id
there. Besides that, you could even merge your node and your hierarchy table.

Dont forget to create indexes on the primary and foreign keys.

nospamno (AT) nospam (DOT) com (Hitch) wrote in message news:<1g5te0n.1q86ordkt1x8aN%nospamno (AT) nospam (DOT) com>...
Dont know if it is exactly what you are looking for but here is one
attempt:

http://fungus.teststation.com/~jon/t...eeHandling.htm

It describes a way to add "crossgroups" to an hierarchical
organisation. Usage could be to represent projects spanning over parts
of an organisation, or to save results of queries, for later
processing, etc.

I have not tried the link extension with a large dataset, but I
suspect (especially the delete part) to be rather timeconsuming.


HTH
/Lennart

Boris Stumm <stumm (AT) informatik (DOT) uni-kl.de> wrote:

Yes, I could do that but it seems to solve nothing about speed of
search. Remember, that one can search the whole "user-root" directory,
how does our quety looks then? Its terrible.

The only solution that seems to be fast is to create another table, that
contains all node_ids,their childs and all offsprings (like transition
closure,but only for a "user-root" direcotory) so we can have fast
"user-root-search" by joining that table and nodes table.

Anyway, it can be awful to maintain (and slow) if we have many users.

Cheers
H

hi,

Hmm...it seems that what wrote earlier is similiar.
Simply, in situation described in that document, you make a modified
copy of "hierarchy description (transition closure contents)" for each
link. That can take a lot of place and can be slow.

---------------
The only solution that seems to be fast is to create another table, that
contains all node_ids,their childs and all offsprings (like transition
closure,but only for a "user-root" direcotory) so we can have fast
"user-root-search" by joining that table and nodes table.
----------------

Bests
H

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 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;
-- the top column is not needed in the final answer
END WHILE;
-- SELECT node, lft, rgt FROM Stack;
END;

I have a book on TREES & HIERARCHIES IN SQL coming out in April 2004.

Build a tree structure that references the Nodes,as usual. Next,
create aa alias table to relate users and neither the nodes or the
location in the tree structure (not sure about your rules here).

The mistake most people make is putting both nodes and edges (tree
structure) in the same table.

--CELKO-- <joe.celko (AT) northface (DOT) edu> wrote:

Actually, I converte adj to sets
using the folllowing function by me (pgplsql)

CREATE OR REPLACE FUNCTION convert_to_nested_set(int,int) RETURNS int AS
'
DECLARE
fid ALIAS FOR $1;
leftval ALIAS FOR $2;
rec RECORD;
size int;
BEGIN

size:=1;

FOR rec IN SELECT id FROM folders WHERE nullequ(parent_id,fid)
LOOP
size:=size+convert_to_nested_set(rec.id,leftval+si ze);
size:=size+1; -- +1 zeby nastepne lft nie bylo jak
poprzednie sasiadujace rgt
END LOOP;

IF fid IS NOT NULL THEN
UPDATE folders SET lft=leftval,rgt=leftval+size WHERE
id=fid;
END IF;

RETURN size;

END;
'
LANGUAGE plpgsql;

I preferer standard "transition closure" as elements of hierarchies
must be dynamically updateable,and as you know NS are good for rather
"stable" hierarchies. Anyway, the question wasn't about how to do
hierarchies, but how to do aliasing within hierarchies. With transition
closure model solutions are not very efficient unfortunately, but
hopefully i need only part of it.

Cheers
Hitch

nospamno (AT) nospam (DOT) com (Hitch) wrote in message news:<1g5wugy.1v9gjtg1yx2d8cN%nospamno (AT) nospam (DOT) com>...
[...]

Delete link looks like a lot of pain, insert looks like perhaps,
loopup should (almost) be as fast as in a strict tree.


Hmmm, but for me a union would feel more natural, something like:

table tree ( id , parentid )
table tree_tc ( id, ancestorid )
table userroot ( id , parentid )
table userroot_tc ( id, ancestorid )

and then search the subtree(s) like:

select id from tree_tc where ancestorid = ?
union all
select id from userroot_tc where ancestorid = ?

A join would either have to be outer, or you have to force every user
to have a user_root structure. But perhaps thats not a problem.


Kind regards
/Lennart