Advanced graph SQL question

I have a directed graph described by A -> B meaning that there exists a connection, always weight 1, FROM A TO B. The problem I want to solve (This is not an academic project) is how can I tell how many common connections there are between two nodes.

To say that in terms of A and B.

There are 2 things that need to get done,

* To look at all of my, B, links coming in (All A's to some B)

* Count how many common A's out of all My, Original B, A's.

I do not know if that makes sense but I'll show you how far I have come.

* First point.

SELECT A
FROM graph
WHERE B='myid';

As most can tell, part 1 is a very simple question. Part 2 is where things get tricky.

I have been able to get all the A's with at least 1 connection or more similar.

Second point.

SELECT G.A, count( G2.A ) AS common
FROM graph AS G2
JOIN (
开发者_开发百科    SELECT A, B
    FROM graph
    WHERE B = 'myid'
) AS G ON G.A = G2.B

So the second point is close because it will return all common links, but it will not return all links which have no common links. Is there a way to get that too?

---

There is still confusion: Ill try to draw a picture... with words.

Here is the table.

A, B
-----   
2, 1  
3, 1  
2, 3

If I wanted to see how many common links from all incoming links into NODE 1 I should see

A, count
---------  
2, 1 // This is for 2's connection to 3.  
3, 0  

With the current SQL statement I have I see this.

A, count
---------  
2, 1 // This is for 2's connection to 3.  

---

Instead of using a subquery, I would just use JOINs:

SELECT
    N1.A,
    COUNT(N3.A)
FROM
    Nodes N1
INNER JOIN Nodes N2 ON
    N2.B = 'myid'
LEFT OUTER JOIN Nodes N3 ON
    N3.A = N1.A AND
    N3.B = N2.A
WHERE
    N1.B = 'myid'
GROUP BY
    N1.A

This gives the answer that you are expecting in your example. You may want to test it out on a wider sample.

You should also do performance testing on this if your data set is of any significant size.