質問

I have a table that looks like:

 a | b
---+---
 1 | a
 2 | a
 3 | a

 1 | b
 3 | b

 2 | c
 3 | c

It represents many-to-many relatons a<->b. I'd like to get all existing relations a<->count(b)<->a, like:

 a1 | a2 | count
----+----+-------
 1  |  2 |     1         #1<->a<->2
 1  |  3 |     2         #1<->(a,b)<->3

 2  |  1 |     1         #duplicate for 1<->a<->2
 2  |  3 |     2         #2<->(a,c)<->3

 3  |  1 |     2         #duplicate for 1<->(a,b)<->3 
 3  |  1 |     2         #duplicate for 2<->(a,c)<->3

I've managed it yet for single a, but can't figure out how to cycle through all:

SELECT
 '1' AS a1,
 t1.a AS a2,COUNT(t1.b)
FROM 
 a_b t1
INNER JOIN(
  SELECT
   b
  FROM a_b
  WHERE
   a = '1'
  ) t2
ON
 t1.b = t2.b
WHERE t1.a != '1'
GROUP BY t1.a
ORDER BY t1.a;

 a1 | a2 | count
----+----+-------
 1  |  2 |     1
 1  |  3 |     2

Is it achievable without cross joining a_b on itself or looping through external script?

Here is SQLFiddle http://www.sqlfiddle.com/#!1/8b53a/1/0

TIA

役に立ちましたか?

解決 2

A little addition to Gordon Linoff's solution: In order to not get the relations double (i.e. 1-3 and 3-1) I added the where clause:

select ab1.a as a1, ab2.a as a2, count(*)
from a_b ab1 join
     a_b ab2
     on ab1.b = ab2.b and ab1.a <> ab2.a
     where ab1.a < ab2.a
group by ab1.a, ab2.a

他のヒント

I think this is a basic "join" query with aggregation:

select ab1.a, ab2.a, count(*)
from a_b ab1 join
     a_b ab2
     on ab1.b = ab2.b and ab1.a <> ab2.a
group by ab1.a, ab2.a
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