list of friend not being display using php and mysql - php

Question

i have a list of friend that i want to show on each profile page where the user is friend with them but the problem is that when i use firebug the system display one id and with a default piture
so i think the error is in the foreach loop can anyone help me ????
php code:
<?php
//***********************Displaying Friend List*************************//
$friendListTitle = "";
$friendList = "";
if($friend_array!="")
{
$friendArray = explode(",", $friend_array);
$friendArray = array_slice($friendArray,0,6);
$friendCount = count($friendArray);
var_dump($friendCount);
$friendListTitle = '<div class="title"> '.$username.'\'s Friends('.$friendCount.')</div>';
//iterating to retrieve what it's needed as values
/*$frnd1 = $friendArray[0];
$frnd2 = $friendArray[1];
/*$frnd3 = $friendArray[2];
$frnd4 = $friendArray[3];
$friendList .='<div style="background-color:"#CCC";>'.$frnd1.'<br />'.$frnd2.'</div>';*/
$i=0;
$friendList .='<div style="background-color:"#CCC"; >';
foreach($friendArray as $key => $value)
{
$i++;
$check_pic = "members/$value/image01.jpg";
if(file_exists($check_pic))
{
$frnd_pic = '<img src = \"$check_pic\" width = \"30px\"; border = \"1\"/>';
}
else
{
$frnd_pic = '<img src = "members/0/image01.jpg" width = \"30px\" border = \"1\"/> ';
}
$sqlName = mysql_query("SELECT first_name, last_name FROM members WHERE user_id= '$value' LIMIT 1") or die(mysql_error());
while($row = mysql_fetch_array($sqlName))
{
$fname = $row['first_name'];
$lname = $row['last_name'];
$friendList = '<div title="'.$fname.' '.$lname.'">'.$frnd_pic.'</div>';
}
}
$friendList .='</div>';
}
?>

Answer 1

Your html table structure is broken:
add closing bracket here <table border = "0" align="center" cellpadding="3", add td tags, put tr tags into loop
also, replace
if($row = mysql_fetch_array($sqlName,MYSQL_ASSOC))
with
while($row = mysql_fetch_array($sqlName,MYSQL_ASSOC))

Answer 2

You have missed one dot. Please look at this
while($row = mysql_fetch_array($sqlName))
{
$fname = $row['first_name'];
$lname = $row['last_name'];
$friendList .= '<div title="'.$fname.' '.$lname.'">'.$frnd_pic.'</div>';
}
because the varibale's name ($friendList) you need to make some string concatenation not simple substitution.
If You use substitution then everythinh what you have collected in $friendList will be lost.
The main problem was in this line:
$friendList .= '<div title="'.$fname.' '.$lname.'">'.$frnd_pic.'</div>';