Recursive query in SQL
Olivier Gaumond
example, if I want to represent a tree in a table with two colums:
Node and Parent. I want a query that returns all the nodes in the
sub-tree of a determined node.
If I simply want the child nodes I could have something like:
SELECT Node WHERE Parent = 1;
If I want a certain level of nodes I could use:
SELECT Node WHERE Parent IN (SELECT Node WHERE Parent = 1);
And I could nest SELECT statements like that for ever, but is there a
simple an elegant way?
I guess it could be done quite simply with a procedural language (like
PL/SQL) however i'm looking for something using standard SQL-DML
Olivier
Alan
times and Joe Celko will post a lengthy discussion in response.
"Olivier Gaumond" <ogau...@yahoo.com> wrote in message
news:b3ae677.02052...@posting.google.com...
--CELKO--
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.html
http://searchdatabase.techtarget.com/tip/1,289483,sid13_gci801943,00.html
Lennart Jonsson
> Is there a way to make a recursive query in standard SQL? For
> example, if I want to represent a tree in a table with two colums:
> Node and Parent. I want a query that returns all the nodes in the
> sub-tree of a determined node.
>
DB2 has a "with ... union all ..." construct that can be used for
recursive queries. See for example:
http://www7b.boulder.ibm.com/dmdd/library/techarticle/0203venigalla/0203venigalla.html
Oracle has a connect by construct (sorry no ref)
/Lennart
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