Wow, I sure went overboard with the "friendship chain" thought. I don't know where I got the idea that was your question.
On Thu, Dec 18, 2014 at 3:46 PM, John McKown <john.archie.mckown@xxxxxxxxx> wrote:
I have a table called friends with a user_id and a friend_id (both of these relate to an id in a users table).For each friend relationship there are two rows. There are currently ONLY reciprocal relationships. So if user ids 1 and 2 are friends there will be two rows (1,2) and (2,1).For 2 arbitrary ids, I need a query to get two pieced of data:* Are the two users friends?* How many friends do the two users have in common.Is there a way to do this with one query? Currently I've only been able to figure out how to do it with two.SELECTEXISTS(SELECT 1FROM friendsWHERE user_id = 166324 AND friend_id = 166325) AS friends,(SELECT COUNT(1)FROM friends f1 JOIN friends f2 ON f1.friend_id = f2.friend_idWHERE f1.user_id = 166324 AND f2.user_id = 166325) AS mutual;I'm wondering if there is a better way to do this using only one query. I've tried a couple of GROUP BY approaches but they haven't worked.This appears, to me, to require a RECURSIVE CTE. Similar to the description on http://www.postgresql.org/docs/9.1/static/queries-with.html towards the bottom, when it goes into avoiding loops on parts which are made up of sub-parts which are themselves sub-parts to other parts. In your case, this would be to eliminate multiple friendship paths which lead to a given person. I.e. A friend of B, friend of C, friend of D, friend of B leading to a recursive loop. In particular, the example:WITH RECURSIVE search_graph(id, link, data, depth, path, cycle) AS ( SELECT g.id, g.link, g.data, 1, ARRAY[g.id], false FROM graph g UNION ALL SELECT g.id, g.link, g.data, sg.depth + 1, path || g.id, g.id = ANY(path) FROM graph g, search_graph sg WHERE g.id = sg.link AND NOT cycle ) SELECT * FROM search_graph;Could be a template for you to start with. Where "id" is the "user_id" and "link" is the "friend_id". You could use that CTE to create a VIEW where "search_graph" is "friends_of_friends". I don't have an exact query for you, sorry. You then use the VIEW to do something like:-- number of friends in common:SELECT COUNT(*) FROM (SELECT friend_id FROM friends_of_friends WHERE user_id = 166324INTERSECTSELECT friend_id FROM friends_of_friends WHERE user_id = 166325)-- Are two people direct friends:SELECT user_id, friend_id FROM friendsWHERE user_id = 16634 AND friend_id = 166325OR user_id = 166325 AND friend_id = 166324;If you want a "transitive" friendship, use the friends_of_friends view instead of the friends table.--While a transcendent vocabulary is laudable, one must be eternally careful so that the calculated objective of communication does not become ensconced in obscurity. In other words, eschew obfuscation.111,111,111 x 111,111,111 = 12,345,678,987,654,321Maranatha! <><
John McKown
While a transcendent vocabulary is laudable, one must be eternally careful so that the calculated objective of communication does not become ensconced in obscurity. In other words, eschew obfuscation.111,111,111 x 111,111,111 = 12,345,678,987,654,321
John McKown