Very Computer

by **Lasse Edsvi** » Wed, 13 Sep 2000 07:15:19

Hello

how is this done....

I have

ID (ID/PK), Parent, Name
1, 0, A
2, 1, B
3,0,C
4,1,D

and if I delete a row that has parent=0 I want to delete those who has
parent=id for the one deleted

e.g

let's say I delete id=1, then I want it to delete id=1 and id=2

but if I delete id=2 I want it to just delete id=2

any clues?
/Lasse

by **Gabor Bod** » Tue, 12 Sep 2000 22:35:27

tjena
i think
'
x=1
delete * from table where ID=x or Parent=x
'
should do it

Quote:> Hello

> how is this done....

> I have

> ID (ID/PK), Parent, Name
> 1, 0, A
> 2, 1, B
> 3,0,C
> 4,1,D

> and if I delete a row that has parent=0 I want to delete those who has
> parent=id for the one deleted

> e.g

> let's say I delete id=1, then I want it to delete id=1 and id=2

> but if I delete id=2 I want it to just delete id=2

> any clues?
> /Lasse

by **Lasse Edsvi** » Wed, 13 Sep 2000 07:42:56

but if parent is NOT 0 for the id I just want to delete that particular one!

> tjena
> i think
> '
> x=1
> delete * from table where ID=x or Parent=x
> '
> should do it

> > Hello

> > how is this done....

> > I have

> > ID (ID/PK), Parent, Name
> > 1, 0, A
> > 2, 1, B
> > 3,0,C
> > 4,1,D

> > and if I delete a row that has parent=0 I want to delete those who has
> > parent=id for the one deleted

> > e.g

> > let's say I delete id=1, then I want it to delete id=1 and id=2

> > but if I delete id=2 I want it to just delete id=2

> > any clues?
> > /Lasse

by **Lasse Edsvi** » Wed, 13 Sep 2000 07:44:34

I need to use either "if" or "case"

> tjena
> i think
> '
> x=1
> delete * from table where ID=x or Parent=x
> '
> should do it

> > Hello

> > how is this done....

> > I have

> > ID (ID/PK), Parent, Name
> > 1, 0, A
> > 2, 1, B
> > 3,0,C
> > 4,1,D

> > and if I delete a row that has parent=0 I want to delete those who has
> > parent=id for the one deleted

> > e.g

> > let's say I delete id=1, then I want it to delete id=1 and id=2

> > but if I delete id=2 I want it to just delete id=2

> > any clues?
> > /Lasse

by **Lasse Edsvi** » Wed, 13 Sep 2000 07:56:34

here's some psuedo-code:

case when parent=0
delete table where parent=1 or id=1
else
delete table where id=1
end

how is it exactly done?

> I need to use either "if" or "case"

> > tjena
> > i think
> > '
> > x=1
> > delete * from table where ID=x or Parent=x
> > '
> > should do it

> > > Hello

> > > how is this done....

> > > I have

> > > ID (ID/PK), Parent, Name
> > > 1, 0, A
> > > 2, 1, B
> > > 3,0,C
> > > 4,1,D

> > > and if I delete a row that has parent=0 I want to delete those who has
> > > parent=id for the one deleted

> > > e.g

> > > let's say I delete id=1, then I want it to delete id=1 and id=2

> > > but if I delete id=2 I want it to just delete id=2

> > > any clues?
> > > /Lasse

by **Gabor Bod** » Tue, 12 Sep 2000 23:06:38

oh sorry.
maybe like this:
dim sql
rs.open "select * from table where
whateverfieldyoursearchingfor='whatevervalue'",yourconnectionstring,adopen?,
adlock?,adcmdtext"
if rs("parent")=0 then
rs.delete
end if

rs is a recordset.
hope this helps

> I need to use either "if" or "case"

> > tjena
> > i think
> > '
> > x=1
> > delete * from table where ID=x or Parent=x
> > '
> > should do it

> > > Hello

> > > how is this done....

> > > I have

> > > ID (ID/PK), Parent, Name
> > > 1, 0, A
> > > 2, 1, B
> > > 3,0,C
> > > 4,1,D

> > > and if I delete a row that has parent=0 I want to delete those who has
> > > parent=id for the one deleted

> > > e.g

> > > let's say I delete id=1, then I want it to delete id=1 and id=2

> > > but if I delete id=2 I want it to just delete id=2

> > > any clues?
> > > /Lasse

by **Gabor Bod** » Tue, 12 Sep 2000 23:16:20

How do you mean? If parent=0 then you want to delete where parent=1 and
id=1?
in that case, it should be

rs.open "select * from table where ???=???",???,???,???,???
if rs("parent")=0
then
rs.close
rs.open "select * from table where parent=1 and id=1",???,???,???,???
rs.delete
else
rs.open "select * from table where id=1",???,???,???,???
rs.delete
end if

> here's some psuedo-code:

> case when parent=0
> delete table where parent=1 or id=1
> else
> delete table where id=1
> end

> how is it exactly done?

> > I need to use either "if" or "case"

> > > tjena
> > > i think
> > > '
> > > x=1
> > > delete * from table where ID=x or Parent=x
> > > '
> > > should do it

> > > > Hello

> > > > how is this done....

> > > > I have

> > > > ID (ID/PK), Parent, Name
> > > > 1, 0, A
> > > > 2, 1, B
> > > > 3,0,C
> > > > 4,1,D

> > > > and if I delete a row that has parent=0 I want to delete those who
has
> > > > parent=id for the one deleted

> > > > e.g

> > > > let's say I delete id=1, then I want it to delete id=1 and id=2

> > > > but if I delete id=2 I want it to just delete id=2

> > > > any clues?
> > > > /Lasse

by **Joe Celk** » Wed, 13 Sep 2000 00:23:29

>> 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
When posting, inclusion of SQL (CREATE TABLE ..., INSERT ..., etc)
which can be cut and pasted into Query Analyzer is appreciated.

Sent via Deja.com http://www.deja.com/
Before you buy.

by **Lasse Edsvi** » Wed, 13 Sep 2000 10:19:16

Thank you celko :)

I already have that reply saved on my computer :) this time it's only 1
level so no celko-systems required, although I've used some of your theory
and I must say it really works great :)

Best regards
/Lasse

Joe Celko wrote:
> >> 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
> When posting, inclusion of SQL (CREATE TABLE ..., INSERT ..., etc)
> which can be cut and pasted into Query Analyzer is

...

read more »

by **Itzik Ben-Ga** » Wed, 13 Sep 2000 02:54:01

DELETE FROM T1

OR parent = (SELECT id
FROM T1

--
BG

Hi-Tech College, Israel
http://sql.hi-tech.co.il

> here's some psuedo-code:

> case when parent=0
> delete table where parent=1 or id=1
> else
> delete table where id=1
> end

> how is it exactly done?

> > I need to use either "if" or "case"

> > > tjena
> > > i think
> > > '
> > > x=1
> > > delete * from table where ID=x or Parent=x
> > > '
> > > should do it

> > > > Hello

> > > > how is this done....

> > > > I have

> > > > ID (ID/PK), Parent, Name
> > > > 1, 0, A
> > > > 2, 1, B
> > > > 3,0,C
> > > > 4,1,D

> > > > and if I delete a row that has parent=0 I want to delete those who
has
> > > > parent=id for the one deleted

> > > > e.g

> > > > let's say I delete id=1, then I want it to delete id=1 and id=2

> > > > but if I delete id=2 I want it to just delete id=2

> > > > any clues?
> > > > /Lasse

by **Ivan Arjentinsk** » Wed, 13 Sep 2000 02:44:20

Lasse,

Seems like you are totally confused with the tree implementation.

The pseudo algorithm you posted will not do the job you expect.

Just pick any well defined tree handling algorithm and use it.
You can try the thread:
"Self referencing table vs. tree traversal method" from Hadjeboopday,
originated on d/m/yyyy-01.09.2000.

--
Ivan Arjentinski
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"Joe Celko" <71062.1...@compuserve.com> wrote in message

news:8pitcm$h7n$1@nnrp1.deja.com...

> >> and if I delete a row that has parent=0 I want to delete those who
> has parent=id for the one deleted <<