Re: ORDER BY with LTREE - Mailing list pgsql-sql

From Michael Fuhr
Subject Re: ORDER BY with LTREE
Date
Msg-id 20060221063133.GA84642@winnie.fuhr.org
Whole thread Raw
In response to ORDER BY with LTREE  (Axel Straschil <axel@straschil.com>)
Responses Re: ORDER BY with LTREE
List pgsql-sql
On Mon, Feb 20, 2006 at 11:15:43AM +0100, Axel Straschil wrote:
> I have a table with a ltree and acaption "CREATE TABLE (tree LTREE, 
> caption TEXT)", I can't store the data of caption in the ltree, becouse of 
> the limitation of the ltree labels. I use the ltree to give the captions 
> a direcory like hirachy, wenn a new caption is createt, the customer can 
> decide which is the parent caption.
> 
> My Problem now ist, that I want to show in a list-box all the captions 
> ordered by ltree, but in the same hirachy by the caption.

Maybe somebody can think of a better way, but as I mentioned in my
previous post you could create a custom operator to use in the ORDER
BY clause.  The operator's function would receive two rows: if the
rows have a common parent in the hierarchy then the function could
compare the rows' labels; otherwise it could query for other rows
and perform appropriate comparisons.  With a large table those queries
could make ordering expensive; if anybody has a better suggestion
then hopefully they'll mention it.

Below is a simple example using the data from your original message
and giving the results you requested.  It's intended to be a starting
point, not a final solution.  I have only a little experience with
ltree and with using custom operators in this manner, so corrections
and improvements are welcome.

CREATE TABLE foo (tree ltree, caption text);

INSERT INTO foo VALUES ('root.1', 'z');
INSERT INTO foo VALUES ('root.2', 'c');
INSERT INTO foo VALUES ('root.2.1', 'a');
INSERT INTO foo VALUES ('root.2.2', 'b');
INSERT INTO foo VALUES ('root.3', 'i');
INSERT INTO foo VALUES ('root.4', 'f');
INSERT INTO foo VALUES ('root.4.1', 'k');
INSERT INTO foo VALUES ('root.4.2', 'c');

CREATE FUNCTION foo_lt(foo, foo) RETURNS boolean AS $$
DECLARE   level  integer;   row1   foo%ROWTYPE;   row2   foo%ROWTYPE;
BEGIN   IF subpath($1.tree, 0, -1) = subpath($2.tree, 0, -1) THEN       RETURN $1.caption < $2.caption;   ELSIF $1.tree
@>$2.tree THEN       RETURN true;   ELSE       level := nlevel(lca($1.tree, $2.tree)) + 1;       SELECT INTO row1 *
FROMfoo WHERE tree = subpath($1.tree, 0, level);       SELECT INTO row2 * FROM foo WHERE tree = subpath($2.tree, 0,
level);      RETURN row1.caption < row2.caption;   END IF;
 
END;
$$ LANGUAGE plpgsql STABLE STRICT;

CREATE OPERATOR < (   PROCEDURE = foo_lt,   LEFTARG   = foo,   RIGHTARG  = foo
);

SELECT * FROM foo ORDER BY foo USING <;  tree   | caption 
----------+---------root.2   | croot.2.1 | aroot.2.2 | broot.4   | froot.4.2 | croot.4.1 | kroot.3   | iroot.1   | z
(8 rows)

-- 
Michael Fuhr


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