Thread
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Recursive select
Jason Kwok <jason@newhonest.com> — 2001-10-26T06:57:18Z
I have a table : MyID parentID ================ 5 6 6 7 3 13 7 3 Is there any simple select statement that can get all rows with MyID = 5 and all its parents? That means 5's parenet is 6, 6's parent is 7, 7's parent is 3..... Jason --- Virus Free Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.286 / Virus Database: 152 - Release Date: 2001/10/9
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Re: Recursive select
Esteban Gutierrez Abarzua <esgutier@sauce.chillan.ubiobio.cl> — 2001-10-29T18:57:37Z
On Fri, 26 Oct 2001, Jason Kwok wrote: > I have a table : > > MyID parentID > ================ > 5 6 > 6 7 > 3 13 > 7 3 > > Is there any simple select statement that can get all rows with MyID = 5 and > all its parents? > That means 5's parenet is 6, 6's parent is 7, 7's parent is 3..... > > > Jason > I think that this is not possible. You may do it using sql embedded or jdbc! bye.
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Re: Recursive select
Keith Gray <keith@heart.com.au> — 2001-10-29T22:09:31Z
Jason Kwok wrote: > > I have a table : > > MyID parentID > ================ > 5 6 > 6 7 > 3 13 > 7 3 > > Is there any simple select statement that can get all rows with MyID = 5 and > all its parents? > That means 5's parenet is 6, 6's parent is 7, 7's parent is 3..... > I had a similar problem a couple of weeks back. The only way I could get around it was "programmatically". This is an iterative (loop) problem. Because I am writing for several brands/flavours of database I solved it in VB, but some solutions (for postgres only) are offered using functions or triggers in PgSQL. -- Keith Gray Technical Development Manager Heart Consulting Services P/L mailto:keith@heart.com.au
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Re: Recursive select
Roberto Mello <rmello@cc.usu.edu> — 2001-10-29T22:47:54Z
On Fri, Oct 26, 2001 at 02:57:18PM +0800, Jason Kwok wrote: > I have a table : > > MyID parentID > ================ > 5 6 > 6 7 > 3 13 > 7 3 > > Is there any simple select statement that can get all rows with MyID = 5 and > all its parents? Dan Wickstrom, of the OpenACS.org team, implemented a bunch of PL/pgSQL functions to do what you're after. It simulates the CONNECT BY behaviour in Oracle. OpenACS 4 is GPL'd, so you can look at the source. I need to extract the relevant sections and post them in the PostgreSQL Cookbook (actually, I think there are a couple tree functions in there - http://www.brasileiro.net/postgres/) -Roberto -- +----| http://fslc.usu.edu USU Free Software & GNU/Linux Club |------+ Roberto Mello - Computer Science, USU - http://www.brasileiro.net http://www.sdl.usu.edu - Space Dynamics Lab, Developer Always forgive your enemies. They hate that! -
Re: Recursive select
--CELKO-- <71062.1056@compuserve.com> — 2001-11-01T01:14:08Z
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: 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. DECLARE Out_Tree CURSOR FOR SELECT P1.lft, 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) -
Re: Recursive select
knut.suebert@web.de — 2001-11-04T17:03:06Z
--CELKO-- schrieb: > Another way of representing trees is to show them as nested sets. Good evening, that's what I needed! To limit the result to entries below one node, I'd use something like SELECT P1.lft, COUNT(P2.emp) AS indentation, P1.emp FROM Personnel AS P1, Personnel AS P2 WHERE P1.lft BETWEEN P2.lft AND P2.rgt AND p1.lft>(SELECT lft FROM personnel WHERE emp='Chuck') AND p1.rgt<(SELECT rgt FROM personnel WHERE emp='Chuck') GROUP BY P1.emp, p1.lft ORDER BY P1.lft; lft | indentation | emp -----+-------------+------------ 5 | 3 | Donna 7 | 3 | Eddie 9 | 3 | Fred (3 rows) for emp='Albert' it returns lft | indentation | emp -----+-------------+------------ 2 | 2 | Bert 4 | 2 | Chuck 5 | 3 | Donna 7 | 3 | Eddie 9 | 3 | Fred (5 rows) My question is, how to limit this result to (Albert's indentation)+1? Thanks, Knut Sübert -
Support for "nested sets" and PHP (Re: Recursive select)
knut.suebert@web.de — 2002-03-08T18:38:11Z
--CELKO-- schrieb: > The usual example of a tree structure in SQL books is called an > adjacency list model and it looks like this: > Another way of representing trees is to show them as nested sets. Hello, I thought a while about those nested sets to avoid recursive selects. Made a few functions. Some of the expensive stuff seems to be solved by introducing a 3rd column called "lvl". If you are interested in, you can find it at http://www.net-one.de/~ks/WOoK/ by taking a look at "The PostrgreSQL side". Celko's original text is also there. Some stuff may be interesting for using PHP to edit PostgreSQL's tables. But that part is incomplete, yet. I'd be happy to get your thoughts to improve the handling (and my understanding) of "nested sets". Bye, Knut Sübert