Oracle FAQ | Your Portal to the Oracle Knowledge Grid |
Home -> Community -> Usenet -> c.d.o.misc -> Re: about hierarchical query
>> how can I write an hierarchical query to view all level of my tree,
but only the branches end with leaves that satisfy some conditions? <<
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
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
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:
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 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 = :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
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.
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;
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;
-- SELECT node, lft, rgt FROM Stack; -- the top column is not needed in the final answer -- move stack contents to new tree tableEND; I have a book on TREES & HIERARCHIES IN SQL which you can get at Amazon.com right now.
The Oracle proprietary CONNECT BY clause is a non-relational hidden cursor which is both slow and weak. Received on Sun Nov 21 2004 - 11:49:24 CST