Complex SQL problem

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I have a table with I call a Keyword table. There are 3 columns: ID, Keyword and Parent. The ID is an autonumber and the parent points to the ID of another row. I can thus store a hierarchy with an unlimited number of levels. The top level has parent = 0
I can drill down the hierarchy by running
SELECT * from tblKeyword where <Parent> = I

What I want to do is show an indication whether there are any more levels to drill down to. So when I run the above SQL, I also need to know for each row returned if there are any rows with a parent matching the ID

Anyone know some SQL that will do it efficiently?
 
Unfortunately, I know of know way to get SQL to do
recursion.

Your question is essentially same as the old
Employee/Supervisor relationship. It is easy to show an
employee's supervisor, but keep going up the chain and it
gets harder.

You can do it for one, but how many levels do you want to
go down?

This SQL should (air code) do 1 level.

Select tblKeyword.*,tblKeyword_1.* from tblKeyword Inner
Join tblKeyword as tblKeyword_1 on tblKeyword.Parent =
tblKeyword_1.ID

Chris

-----Original Message-----
I have a table with I call a Keyword table. There are 3
columns: ID, Keyword and Parent. The ID is an autonumber
and the parent points to the ID of another row. I can thus
store a hierarchy with an unlimited number of levels. The
top level has parent = 0.
I can drill down the hierarchy by running:
SELECT * from tblKeyword where <Parent> = ID

What I want to do is show an indication whether there are
any more levels to drill down to. So when I run the above
SQL, I also need to know for each row returned if there
are any rows with a parent matching the ID.
 
Look at

http://allenbrowne.com/AppInventory.html

And also an old post by Joe Celko might give you some ideas:

Quote:

The usual example of a tree structure in SQL books 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, 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 emps are adjacent to
each other.

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 first table is denormalized in several ways. We are modeling both
the OrgChart 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 OrgChart 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 OrgChart),
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 emps with ovals, then nest
subordinate ovals inside each other. The root will be the largest oval
and will contain every other emp. The leaf emps 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 emp, 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 emp 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 emp, he puts a number
in the cell on the side that he is visiting and increments his counter.
Each emp 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 emps 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.

DECLARE Out_Tree CURSOR FOR
SELECT O1.lft, 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.emp
ORDER BY O1.lft;

5. The nested set model has an implied ordering of siblings which
theadjacency 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 right_most_sibling INTEGER;

SET right_most_sibling
= (SELECT rgt
FROM Frammis
WHERE part = 'G');
UPDATE Frammis
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 Frammis (part, qty, wgt, lft, rgt)
VALUES ('G1', 3, 4, parent, (parent + 1));
COMMIT WORK;
END;

The idea is to spread the lft and 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, P.emp
FROM OrgChart AS P
LEFT OUTER JOIN
OrgChart AS B
ON B.lft
= (SELECT MAX(lft)
FROM OrgChart AS S
WHERE P.lft > S.lft
AND P.lft < S.rgt);

For details, see the chapter in my book JOE CELKO'S SQL FOR SMARTIES
(Morgan-Kaufmann, 1999, second edition)

http://searchdatabase.techtarget.com/tip/1,289483,sid13_gci537290,00.htm
l

http://searchdatabase.techtarget.com/tip/1,289483,sid13_gci801943,00.htm
l


--CELKO--


--
Regards,

Adrian Jansen
J & K MicroSystems
Microcomputer solutions for industrial control
 
David,

Although I haven't exhaustively tested this (due to lack of time), on
initial testing it seems to work:

SELECT K.*, Tmp3.AnyMore As HowManyMore
FROM tblKeyword AS K
INNER JOIN (SELECT Tmp2.Parent, Count(Tmp1.ID) As AnyMore
FROM tblKeyword As Tmp1
INNER JOIN tblKeyword As Tmp2
ON Tmp2.Parent = Tmp1.ID
GROUP BY Tmp2.Parent) AS Tmp3
ON K.ID = Tmp3.Parent
WHERE K.Parent = [Enter an ID]

One thing I did notice was that it also counts sub-records. For example, if
we have the following records, the query also returns a count of RecordD
(and any others whose Parent value lead upwards to RecordB. This may or may
not be what you want.
RecordA.ID <-- RecordB.Parent
RecordB.ID <-- RecordC.Parent
RecordC.ID <-- RecordD.Parent

Regards,
Graham R Seach
Microsoft Access MVP
Sydney, Australia

David said:
I have a table with I call a Keyword table. There are 3 columns: ID,
Keyword and Parent. The ID is an autonumber and the parent points to the ID
of another row. I can thus store a hierarchy with an unlimited number of
levels. The top level has parent = 0.
I can drill down the hierarchy by running:
SELECT * from tblKeyword where <Parent> = ID

What I want to do is show an indication whether there are any more levels
to drill down to. So when I run the above SQL, I also need to know for each
row returned if there are any rows with a parent matching the ID.
 
Okay, that is cool. I haven't had to do this before, but
will save this for future reference.


Thanks, and ignore my earlier post.

Chris
 
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