hi everyone!
i`m trying to display a friend of a friend from a friend table,but i dont know how to do it.
i have a table called friends,which has id,friendid and userid fields.
Now i want if a member loggin, to be displayed a list of friends of his friends which are not his friends.
i hope u u got me guys!i`m hopping to hear frm You.
below is what i tried to but it gave me infinite datas.

<?php
	   //Connect to the database server
	
    include"config.php";
    $user=$_SESSION['user'];
    $logged_id=$_SESSION['id'];
	//select all your friends
	$select=mysql_query("SELECT *FROM friends WHERE userid='$logged_id' ORDER BY RAND() LIMIT 2");
	$count_friends=mysql_num_rows($select);
	 if($count_friends>0){
	   while($r=mysql_fetch_array($select)){
	      $friendid=$r['friendid'];
		   //select friends of a friend except yourself and which ar not ur friends
		    $sqr=mysql_query("SELECT *FROM friends WHERE ((userid='$friendid') and (((userid !='$logged_id') or (friendid !='$friendid')) or (userid !='$friendid') or (friendid !='$logged_id')))");
			  while($raw=mysql_fetch_array($sqr)){
			      $friendoffriend=$raw['friendid'];
				  $_query = mysql_query("SELECT * FROM profile WHERE id = '$friendoffriend'");
                  $_row = mysql_fetch_array($_query);
                  $name=$_row['names'];
				  echo"$name fomt";
				  
				  
			  }
			 
	   
	}
	 }else{
	 
	 echo"You must add friends";
	   
	 } 
?>

What do you mean when you say it is giving you infinite results? It is not limiting your result set to 2 ?

wELL IT SEEMS LIKE AFTER I PUT THE
WHILE LOOP TWO TIMES ,IT IS MAKING MY SCRIPT TO LOOP MANY TIMES AND GIVE LOTS OF RESULTS.

CAN YOU SUGGEST ME A BETTER WAY OF DOING THIS PLEASE???

I'm a bit confused about your second query but I'll take your word for it that it is doing what it's supposed to. Set a limit on your second query just like you did on the first, it will limit your result set to just 2.

$sqr=mysql_query("SELECT *FROM friends WHERE ((userid='$friendid') and (((userid !='$logged_id') or (friendid !='$friendid')) or (userid !='$friendid') or (friendid !='$logged_id'))) ORDER BY RAND() LIMIT 2");

I'm not sure what criteria your query is looking for but it does look like a lot of it could be cleaned up an simplified by using a JOIN or a subquery.

sO U MEAN THE WHOLE CODE WILL BE LIKE THIS

include"config.php";
    $user=$_SESSION['user'];
    $logged_id=$_SESSION['id'];
$sqr=mysql_query("SELECT *FROM friends WHERE ((userid='$friendid') and (((userid !='$logged_id') or (friendid !='$friendid')) or (userid !='$friendid') or (friendid !='$logged_id'))) ORDER BY RAND() LIMIT 2");

AND HERE THE WHILE LOOP AND DISPLAY ???

Well u said u a bit confused of my code,thats true.!because i wrote that as my first idea And i`m asking You to suggest a how to do this.
HERE IS THE WHOLE TASK I`M SUPPOSED TO DO.
Lets say u login in my a web,and you have friends.
Your friends also have got friends which some of them are not Your friends.
so i want to Display those friends of your friends which are not Your friends.

Structure of Friends Table is
id,friendid and userid (userid is your id after u login,which i store it in sessions.)

I HOPE FROM HERE yOU CAN HELP ME,HOW TO CODE THIS ISSUE.

$sqr=mysql_query("SELECT friendid FROM friends WHERE userid IN (SELECT friendid FROM friends WHERE userid='$logged_id') AND friendid NOT IN (SELECT friendid FROM friends WHERE user_id='$logged_id') ORDER BY RAND() LIMIT 2") or die(mysql_error());

// Do While Loop here

You can try that and see what you get. It may have to be tweaked a little bit.

Thank You,i`m trying it and i`ll update you as soon as possible.

@CFROG,Thank You very..the code works well.
except one thing,i also see myself because i`m also his friend.
I guess it need a little fix to hide myself($logged_id) from the list.

But it is such wonderful, i real appreciate man.

Like I said, it might have to be tweaked a little. I'm sure you can figure that part out on your own.

Glad it worked for you.

Sure! i will try ma self to fix the remaining part.
Thank You again.

Don't over-complicate it, it's an extremely easy fix. Just think it through.

Let me know if you run into a problem.

well here is how i solved it,i didn`t use sql as i`m not Good with Them., SO I JUST CHECKED WIT PHP

$logged_id=$_SESSION['id'];
	$sqr=mysql_query("SELECT friendid FROM friends WHERE userid IN (SELECT friendid FROM friends WHERE userid='$logged_id') AND friendid NOT IN (SELECT friendid FROM friends WHERE userid='$logged_id') ORDER BY RAND() LIMIT 10") or die(mysql_error());
	$countfriends=mysql_num_rows($sqr);
if($countfriends>0){

While($row=mysql_fetch_array($sqr)){
	     $friendidof=$row['friendid'];
               //CHECK IF IS THE LOGGED  ID
		 if($friendidof!=$logged_id){ 

            DISPAY RESULT
 }

}
}else{

echo"I SUGGEST YOU SHOULD ADD FRIENDS";

}

I had something slightly different in mind but it looks like you worked it out just fine either way. I'm glad you got it working the way you want.

yeah man it working Thank You.

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