Re: ORDER BY with exception - Mailing list pgsql-general
From | Michael Glaesemann |
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Subject | Re: ORDER BY with exception |
Date | |
Msg-id | 3C976E5B-7A52-4756-9B22-07C6CE3C013A@seespotcode.net Whole thread Raw |
In response to | ORDER BY with exception (brian <brian@zijn-digital.com>) |
Responses |
Re: ORDER BY with exception
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List | pgsql-general |
On Jun 21, 2007, at 17:35 , brian wrote: > I have a lookup table with a bunch of disciplines: To answer your ordering question first: SELECT id, name FROM discipline ORDER BY name = 'other' , name; id | name ----+--------------------- 8 | community 4 | dance 5 | film and television 9 | fine craft 7 | media arts 3 | music 6 | theatre 2 | visual arts 1 | writing 10 | other (10 rows) This relies on the fact that FALSE orders before TRUE. I don't always remember which way, so I often have to rewrite it using <> or = to get the behavior I want. > and a function that returns each discipline name along with the > total number of records in another table (showcase) that are > related to each discipline. Each showcase entry may have 0 or more > items (showcase_item) related to it, so ones that have no items are > disregarded here. Also, only showcases that have been accepted > should be counted. > > First, here's the working function: I don't think you really need to use a function for this. I believe you should be able to do this all in one SQL statement, something like (if I've understood your query and intent correctly): SELECT discipline.name, COUNT(showcase_id) AS total FROM discipline LEFT JOIN ( SELECT DISTINCT discipline_id, showcase.id as showcase_id FROM showcase JOIN showcase_item on (showcase.id = showcase_id) WHERE accepted) AS accepted_showcases ON (discipline.id = discipline_id) GROUP BY discipline.name ORDER BY discipline.name = 'other' , discipline.name; name | total ---------------------+------- community | 0 dance | 0 film and television | 0 fine craft | 0 media arts | 0 music | 0 theatre | 0 visual arts | 1 writing | 2 other | 0 (10 rows) This should give you the total number of showcases that have been accepted for each discipline. (DDL and data below.) As a general rule, it's generally better to let the server handle the data in sets (i.e., tables) as much as possible rather than using procedural code. Hope this helps. Michael Glaesemann grzm seespotcode net CREATE TABLE discipline ( id INTEGER NOT NULL UNIQUE , name TEXT PRIMARY KEY ); INSERT INTO discipline (id, name) VALUES (1, 'writing') , (2, 'visual arts') , (3, 'music') , (4, 'dance') , (5, 'film and television') , (6, 'theatre') , (7, 'media arts') , (8, 'community') , (9, 'fine craft') , (10, 'other'); SELECT * FROM discipline ORDER BY name; SELECT * FROM discipline ORDER BY name = 'other', name; CREATE TABLE showcase ( id INTEGER NOT NULL UNIQUE , name TEXT PRIMARY KEY , discipline_id INTEGER NOT NULL REFERENCES discipline(id) , accepted BOOLEAN NOT NULL ); INSERT INTO showcase (id, name, discipline_id, accepted) VALUES (1, 'foo', 1, true) , (2, 'bar', 2, true) , (3, 'baz', 1, true) , (4, 'quux', 1, false) , (5, 'blurfl', 2, false); CREATE TABLE showcase_item ( id INTEGER NOT NULL UNIQUE , description TEXT NOT NULL , showcase_id INTEGER NOT NULL REFERENCES showcase (id) , PRIMARY KEY (description, showcase_id) ); INSERT INTO showcase_item (id, description, showcase_id) VALUES (1, 'a', 1) , (2, 'b', 1) , (3, 'c', 1) , (4, 'd', 2) , (5, 'e', 2) , (6, 'f', 2) , (7, 'g', 3) , (8, 'h', 3) , (9, 'i', 4) , (10, 'j', 5); SELECT * FROM showcase; id | name | discipline_id | accepted ----+--------+---------------+---------- 1 | foo | 1 | t 2 | bar | 2 | t 3 | baz | 1 | t 4 | quux | 1 | f 5 | blurfl | 2 | f (5 rows) SELECT * FROM showcase JOIN showcase_item ON (showcase.id = showcase_id); id | name | discipline_id | accepted | id | description | showcase_id ----+--------+---------------+----------+----+------------- +------------- 1 | foo | 1 | t | 1 | a | 1 1 | foo | 1 | t | 2 | b | 1 1 | foo | 1 | t | 3 | c | 1 2 | bar | 2 | t | 4 | d | 2 2 | bar | 2 | t | 5 | e | 2 2 | bar | 2 | t | 6 | f | 2 3 | baz | 1 | t | 7 | g | 3 3 | baz | 1 | t | 8 | h | 3 4 | quux | 1 | f | 9 | i | 4 5 | blurfl | 2 | f | 10 | j | 5 (10 rows)
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