Oracle SQL how to write a sql statement that verifies if user in my network( ie friends or friend of friends)

Question

I have this problem. Given a users table that consists of users\' username in a social network and friends table that contain a user\'s name and a

Answer 1

SELECT  *
FROM    (
        SELECT  username
        FROM    friends
        START WITH
                username = 'myname'
        CONNECT BY
                friendname = PRIOR username
                AND level <= 3
        )
WHERE   username = 'friendname'
        AND rownum = 1

Update the level as necessary: you may search for the third layer friends etc.

If the friendship relationship is symmetric, you should make the following query:

WITH    q AS
        (
        SELECT  username, friendname
        FROM    friends
        UNION ALL
        SELECT  friendname, username
        FROM    friends
        ),
        f AS
        (
        SELECT  friendname, level
        FROM    q
        START WITH
                username = 'Thomas'
        CONNECT BY NOCYCLE
                username = PRIOR friendname
        )
SELECT  *
FROM    f
WHERE   friendname = 'Jo'
        AND rownum = 1

This query can be made much faster if you denormalize your table: store two records per friendship, like this:

CREATE TABLE dual_friends (orestes NOT NULL, pylades NOT NULL, CONSTRAINT pk_dualfriends_op PRIMARY KEY (orestes, pylades)) ORGANIZATION INDEX
AS
SELECT  username, friendname
FROM    friends
UNION ALL
SELECT  friendname, username
        FROM    friends

Then you can just replace the CTE above with the dual_friends:

WITH    f AS
        (
        SELECT  pylades, level
        FROM    dual_friends
        START WITH
                orestes  = 'Thomas'
        CONNECT BY NOCYCLE
                orestes = PRIOR pylades
                AND level <= 3
        )
SELECT  *
FROM    f
WHERE   pylades = 'Jo'
        AND rownum = 1

, which will use the index and be much more efficient, especially if you limit the level to some reasonable value.

Answer 2

You can use connect by