Friend System

pro_se · April 15, 2007

Hi, I recently started an endeavor to create a friend system where it will allow people to add people as friends and I have run into quite a bit of trouble, please help.

..I am STUCK!

I want to display the friends and their pictures as well as links to their page with their friends on that page. But the design of my page only allows 3 thumbnails to be displayed.

I have created the code to display one friend but I cannot get it to display all the friends and limit it to only display 3.

Here is the code that I have managed to create:

<?php function getfriendsforuid($puid) {
$query = "SELECT * FROM `friends` WHERE `passive_id` = '$puid' AND `accpet` = 'y' LIMIT 3";
$result_query = mysql_query($query);
while($fnamelname = mysql_fetch_array($result_query)) {
    $friend = $fnamelname["acting_id"];
$query = "SELECT * FROM `pics` WHERE `uid` = '$friend'";
$result_query = mysql_query($query);
while($fnamelname = mysql_fetch_array($result_query)) {
    $pic = $fnamelname["path"];
}
}
$query2 = "SELECT * FROM `user` WHERE `id` = '$friend'";
$result_query2 = mysql_query($query2);
while($fnamelname2 = mysql_fetch_array($result_query2)) {
    $uname = $fnamelname2["username"];
}
$query3 = "SELECT * FROM `pics` WHERE `uid` = '$friend' AND `default` = 'y'";
$numpics = mysql_query($query3);
$picsfound = mysql_num_rows($numpics);
if ($picsfound >= 1) { 
echo "$pic";
echo "\" onclick=\"window.location = '/profile/$uname'\"";
} else {
echo "npic.jpg\" onclick=\"window.location = '/profile/$uname'\"";
}
} ?>

That only displays one friend.. I was thinking maybe use a while statement to get all the friends but I am not sure of the procedure.

Any and all help is welcomed...

Oh, and here is the function-call:

<?php  echo "<div class=\"friendbar\" style=\"cursor:pointer;\"><img border=\"0\" src=\"/u/s_"; getfriendsforuid($puid); echo "></div>"; ?>

-- just to put it out there...

Thank you..

pro_se · April 16, 2007

gahh.. ANYONE?

I just tried a couple WHILE loops and they were never ending, they repeated over and over again. I cant figure out where I need the loops placed.

anthylon · April 16, 2007

Well I don't know about your tables in DB.

But if your passive_id is numeric field than next query shouldn't work because of apostrophes.

$query = "SELECT * FROM `friends` WHERE `passive_id` = '$puid' AND `accpet` = 'y' LIMIT 3";

Could you show us tables from db. I think you could make it easier with JOIN method or....

pro_se · April 16, 2007

yea, here is the DB Schema

-- phpMyAdmin SQL Dump

-- version 2.8.2.4

-- http://www.phpmyadmin.net

--

-- Host: mysql.cvnetworks.net

-- Generation Time: Apr 15, 2007 at 05:54 PM

-- Server version: 5.0.24

-- PHP Version: 4.4.4

--

-- Database: `notexting`

--

-- --------------------------------------------------------

--

-- Table structure for table `comments`

--

CREATE TABLE `comments` (

`commentid` int(12) NOT NULL auto_increment,

`toid` int(11) NOT NULL,

`fromid` int(11) NOT NULL,

`comment` text NOT NULL,

`datetime` datetime NOT NULL,

PRIMARY KEY (`commentid`)

) ENGINE=MyISAM AUTO_INCREMENT=2 DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ;

-- --------------------------------------------------------

--

-- Table structure for table `friends`

--

CREATE TABLE `friends` (

`friendshipid` int(12) NOT NULL auto_increment,

`acting_id` int(11) NOT NULL,

`passive_id` int(11) NOT NULL,

`date` date NOT NULL,

`accpet` enum('y','n') NOT NULL default 'n',

PRIMARY KEY (`friendshipid`)

) ENGINE=MyISAM AUTO_INCREMENT=5 DEFAULT CHARSET=latin1 AUTO_INCREMENT=5 ;

-- --------------------------------------------------------

--

-- Table structure for table `pics`

--

CREATE TABLE `pics` (

`pid` int(11) NOT NULL auto_increment,

`uid` int(11) NOT NULL,

`path` varchar(50) NOT NULL,

`default` enum('y','n') NOT NULL default 'n',

PRIMARY KEY (`pid`)

) ENGINE=MyISAM AUTO_INCREMENT=3 DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ;

-- --------------------------------------------------------

--

-- Table structure for table `profile`

--

CREATE TABLE `profile` (

`id` int(11) NOT NULL,

`aboutme` text NOT NULL,

`im` varchar(50) default NULL,

PRIMARY KEY (`id`)

) ENGINE=MyISAM DEFAULT CHARSET=latin1;

-- --------------------------------------------------------

--

-- Table structure for table `tags`

--

CREATE TABLE `tags` (

`tagid` int(11) NOT NULL auto_increment,

`uid` int(11) NOT NULL,

`tag` varchar(20) NOT NULL,

PRIMARY KEY (`tagid`)

) ENGINE=MyISAM AUTO_INCREMENT=95 DEFAULT CHARSET=latin1 AUTO_INCREMENT=95 ;

-- --------------------------------------------------------

--

-- Table structure for table `user`

--

CREATE TABLE `user` (

`id` int(11) NOT NULL auto_increment,

`fname` varchar(20) NOT NULL,

`lname` varchar(20) NOT NULL,

`username` varchar(20) NOT NULL,

`password` varchar(32) NOT NULL,

`email` varchar(50) NOT NULL,

`bday` date NOT NULL,

`gender` enum('m','f','o') NOT NULL default 'm',

`orientation` enum('s','b','g','o') NOT NULL default 's',

`location` varchar(50) NOT NULL default 'Fullerton, CA',

`signup_date` date NOT NULL,

`ip` varchar(16) NOT NULL,

`logintimes` int(10) NOT NULL default '0',

`lastlogin` date NOT NULL,

PRIMARY KEY (`id`)

) ENGINE=MyISAM AUTO_INCREMENT=69 DEFAULT CHARSET=latin1 AUTO_INCREMENT=69 ;

pro_se · April 16, 2007

Well I don't know about your tables in DB.

But if your passive_id is numeric field than next query shouldn't work because of apostrophes.
$query = "SELECT * FROM `friends` WHERE `passive_id` = '$puid' AND `accpet` = 'y' LIMIT 3";
Could you show us tables from db. I think you could make it easier with JOIN method or....

Well, the whole thing works, I just need help making the stupid thing display all the rows where the friends id matches whats in the database...

anthylon · April 16, 2007

Is that code all you have? I mean, I don't see that you make connection to database in your getfriendsforuid() function.

You should get error if you want make query without mysql_connect etc.

Explain please.

pro_se · April 16, 2007

Well, I have a 'functions.php' that I include on every page.. Do you want that? I can post it... Its gonna have a lot of irrelevant stuff on it...

pro_se · April 16, 2007

I have made the connection and I have echoed 1 friend. Now what I want to do is echo 3 friends. I was thinking of using a WHILE statement but it did not work when i tried.

anthylon · April 16, 2007

The problem with your code is db link. When you create link on db and execute first query everything is okay. But after you (in first while loop) execute another query, than you lost data in first query. I hope you understand. Basicaly you would need more than one link on database to execute all of your queries.

Please send via email again structure of your data but this time export data too so I can test it and maybe make one query for everything you need.

If I don't answer soon = sleeping already.

pro_se · April 16, 2007

Ok, I sent an email to you..

anthylon · April 16, 2007

Okay, I couldn't read your tables because you made them so "different"....

But, here is what I made:

getfriendsforuid(1);

function getfriendsforuid($puid) {
$cnn1 = mysql_connect('localhost', 'root', 'root');
$cnn2 = mysql_connect('localhost', 'root', 'root');	

$db1 = mysql_select_db('t', $cnn1);
$db2 = mysql_select_db('t', $cnn2);	

$sql1 = "SELECT tbl_users. * , tbl_pictures.path 
		FROM tbl_users
		LEFT JOIN tbl_pictures ON tbl_pictures.id = tbl_users.picture_id
		WHERE tbl_users.id = $puid";

$res1 = mysql_query($sql1, $cnn1);

while($row1 = mysql_fetch_assoc($res1)) {
	echo "<br>User ${row1['fname']} has image: ${row1['path']}<br>His/Her friends are:";
	$id = $row1['id'];
	$sql2 = "SELECT tbl_users. * , tbl_pictures.path
			FROM tbl_users
			LEFT JOIN tbl_friends ON tbl_users.id = tbl_friends.friend_id
			INNER JOIN tbl_pictures ON tbl_users.picture_id = tbl_pictures.id
			WHERE tbl_friends.user_id = $id";
		$res2 = mysql_query($sql2, $cnn2);
		$i = 1;
		while($row2 = mysql_fetch_assoc($res2)) {
			echo "<br>$i. ${row2['fname']} with image ${row2['path']}";
			$i++;		
		}
}
}

Here is new DB exported. So, use it to test:

-- phpMyAdmin SQL Dump
-- version 2.7.0-pl1
-- http://www.phpmyadmin.net
-- 
-- Host (domena): localhost
-- Vrijeme podizanja: 16. Tra 2007. u 05:41
-- Verzija servera: 5.0.11
-- verzija PHP-a: 5.1.0
-- 
-- Baza podataka: `t`
-- 

-- --------------------------------------------------------

-- 
-- Struktura tablice `tbl_friends`
-- 

DROP TABLE IF EXISTS `tbl_friends`;
CREATE TABLE `tbl_friends` (
  `id` int(11) NOT NULL auto_increment,
  `user_id` int(11) NOT NULL,
  `friend_id` int(11) NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ;

-- 
-- Izvoz <i>(dump)</i> podataka tablice `tbl_friends`
-- 

INSERT INTO `tbl_friends` VALUES (1, 1, 2);
INSERT INTO `tbl_friends` VALUES (2, 1, 4);

-- --------------------------------------------------------

-- 
-- Struktura tablice `tbl_pictures`
-- 

DROP TABLE IF EXISTS `tbl_pictures`;
CREATE TABLE `tbl_pictures` (
  `id` int(11) NOT NULL auto_increment,
  `path` varchar(30) NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=5 ;

-- 
-- Izvoz <i>(dump)</i> podataka tablice `tbl_pictures`
-- 

INSERT INTO `tbl_pictures` VALUES (1, 'Image1.jpg');
INSERT INTO `tbl_pictures` VALUES (2, 'Image2.jpg');
INSERT INTO `tbl_pictures` VALUES (3, 'Image3.jpg');
INSERT INTO `tbl_pictures` VALUES (4, 'Image4.jpg');

-- --------------------------------------------------------

-- 
-- Struktura tablice `tbl_users`
-- 

DROP TABLE IF EXISTS `tbl_users`;
CREATE TABLE `tbl_users` (
  `id` int(11) NOT NULL auto_increment,
  `fname` varchar(30) NOT NULL,
  `lname` varchar(30) NOT NULL,
  `picture_id` int(11) NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=5 ;

-- 
-- Izvoz <i>(dump)</i> podataka tablice `tbl_users`
-- 

INSERT INTO `tbl_users` VALUES (1, 'Person1', '', 4);
INSERT INTO `tbl_users` VALUES (2, 'Person2', '', 4);
INSERT INTO `tbl_users` VALUES (3, 'Person3', '', 1);
INSERT INTO `tbl_users` VALUES (4, 'Person4', '', 2);

I tested it and my result for user with ID=1 was:

User Person1 has image: Image4.jpg

His/Her friends are:

1. Person2 with image Image4.jpg

2. Person4 with image Image2.jpg

I hope you can use this to solve your issue.

HeyRay2 · April 16, 2007

Have you tried running this query in a program like PHPMyAdmin to see if you are getting the correct number of result rows?

$query = "SELECT * FROM `friends` WHERE `passive_id` = '$puid' AND `accpet` = 'y' LIMIT 3";

pro_se · April 17, 2007

Well, ok.. That Partially works... My database does not have predefined pictures... The pictures are all user uploaded and the path (or filename) is stored into a row on the database. What you gave me works fine when you are looking up images with a specific ID. I need a way to get the path with just the friends user ID. I dont know how to do all that LEFT JOIN and INNER JOIN stuff so I could not modify it to work for me.. Can you explain to me how that works or can you modify the code you sent me to get the path by looking in the 'pics' table where the ID = PUID and get the path out..

pro_se · April 17, 2007

bump.. anyone?

pro_se · April 17, 2007

okay, I got the stupid thing to have the if/ else thing work.. now.. all I am getting is the same three friends.. all of them are the same.. how can i display all the friends different?

function getfriendsforuid($puid) {

$query = "SELECT * FROM `friends` WHERE `passive_id` = '$puid' AND `accpet` = 'y' LIMIT 3";

$result_query = mysql_query($query);
$numfriends = mysql_num_rows($result_query);
$repeat = 0;
while ($repeat < $numfriends) {
while($fname1lname = mysql_fetch_assoc($result_query)) {
    $friend = $fname1lname["acting_id"];
$query = "SELECT * FROM `pics` WHERE `uid` = '$friend' AND `default` = 'y'";
$result_query = mysql_query($query);
while($fnam2elname = mysql_fetch_assoc($result_query)) {
    $pic = $fnam2elname["path"];
$query2 = "SELECT * FROM `user` WHERE `id` = '$friend'";
$result_query2 = mysql_query($query2);
while($fnamelname2 = mysql_fetch_assoc($result_query2)) {
   $uname = $fnamelname2["username"];
$query3 = "SELECT * FROM `pics` WHERE `uid` = '$friend' AND `default` = 'y'";
$numpics = mysql_query($query3);
$picsfound = mysql_num_rows($numpics);
if ($picsfound >= 1) { 
echo "<div class=\"friendbar\"><a href=\"/profile/$fnamelname2[username]\"><img border=\"0\" src=\"/u/s_$pic\"></a></div>";

}
}
}
}
if ($picsfound == 0) {echo "<div class=\"friendbar\"><a href=\"/profile/$uname\"><img border=\"0\" src=\"/u/s_np.jpg\"></a></div>";}
$repeat++; 
}
}

and a link to my site is: http://notexting.com/profile/cvarma

pro_se · April 18, 2007

ideas?

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