Re: Combining two queries

From: John McKown <john(dot)archie(dot)mckown(at)gmail(dot)com>
To: Robert DiFalco <robert(dot)difalco(at)gmail(dot)com>
Cc: "pgsql-general(at)postgresql(dot)org" <pgsql-general(at)postgresql(dot)org>
Subject: Re: Combining two queries
Date: 2014-12-18 21:46:02
Message-ID: CAAJSdjhYfZ30QhgFdvCC8ujuaET=1R6TqrYRXaNUV-nCuMQObg@mail.gmail.com
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On Thu, Dec 18, 2014 at 2:10 PM, Robert DiFalco <robert(dot)difalco(at)gmail(dot)com>
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.
>
> SELECT
> EXISTS(
> SELECT 1
> FROM friends
> WHERE user_id = 166324 AND friend_id = 166325) AS friends,
> (SELECT COUNT(1)
> FROM friends f1 JOIN friends f2 ON f1.friend_id = f2.friend_id
> WHERE 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 = 166324
INTERSECT
SELECT friend_id FROM friends_of_friends WHERE user_id = 166325
)

-- Are two people direct friends:

SELECT user_id, friend_id FROM friends
WHERE user_id = 16634 AND friend_id = 166325
OR 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 = 12,345,678,987,654,321

Maranatha! <><
John McKown

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