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Home/ Questions/Q 6809775
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T20:07:34+00:00 2026-05-26T20:07:34+00:00

I’m trying to write a friend system that works like this: User A sends

  • 0

I’m trying to write a friend system that works like this:

  1. User A sends friend request to user B.
  2. User B accepts friend request from user A.
  3. Users A and B are now friends.

Here is my database structure:

CREATE TABLE IF NOT EXISTS `users` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `username` varchar(16) NOT NULL,
  `password` char(32) NOT NULL,
  `email` varchar(80) NOT NULL,
  `dname` varchar(24) NOT NULL,
  `profile_img` varchar(255) NOT NULL DEFAULT '/images/default_user.png',
  `created` int(11) NOT NULL,
  `updated` int(11) NOT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `email` (`email`),
  UNIQUE KEY `username` (`username`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=8 ;
CREATE TABLE IF NOT EXISTS `friend_requests` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `user_a` int(11) NOT NULL DEFAULT '0',
  `user_b` int(11) NOT NULL DEFAULT '0',
  `viewed` tinyint(1) NOT NULL DEFAULT '0',
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=8 ;
CREATE TABLE IF NOT EXISTS `friends` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `user_a` int(11) NOT NULL DEFAULT '0',
  `user_b` int(11) NOT NULL DEFAULT '0',
  `friend_type` int(3) NOT NULL DEFAULT '1',
  `friends_since` int(11) NOT NULL DEFAULT '0',
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;

I can list a users friends with “SELECT * FROM friends WHERE user_a = $userid OR user_b = $userID”, but how can I get data such as username or profile_img from the users table?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T20:07:35+00:00Added an answer on May 26, 2026 at 8:07 pm
    SELECT DISTINCT
    a.username,
    a.profile_img
    FROM
    users a
    WHERE
    a.id in (SELECT user_a FROM friends WHERE user_a = $userid OR user_b = $userID)
    and a.id <> $userid
    
    UNION
    
    SELECT
    b.username,
    b.profile_img
    FROM
    users b
    WHERE
    b.id in (SELECT user_b FROM friends WHERE user_a = $userid OR user_b = $userID)
    and b.id <> $userid
    
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