How to display data twice from a mysql table in php - php

Question

Im making a page that will display top ten voters in a table and show the number one voter aswell. I got the top ten table to work but I can't figure out how to get the second part to work. What would I do to get it to work where it would show the top voter in a img tag like this example
img src="https://minotar.net/avatar/TOPVOTERHERE"
First part
<div class="col-md-6">
<table class="table table-striped">
<tr>
<td><strong>Username</strong></td>
<td><strong>Votes</strong></td>
</tr>
<?php
$username2 = "exampleuser";
$password2 = "pass";
$hostname = "127.0.0.1";
$dbhandle = mysql_connect($hostname, $username2 , $password2)
or die("Unable to connect to MySQL");
$selected = mysql_select_db("mc711",$dbhandle)
or die("Could not select database");
$result = mysql_query("SELECT * FROM GALTotals ORDER BY votes DESC");
$count = 0;
while ($row = mysql_fetch_assoc($result)) {
$count++;
if($count>10){
}else{
echo "<tr>";
echo "<td>".$row['IGN']."</td>";
echo "<td>".$row['votes']."</td>";
echo "</tr>";
}
}
?>
</table>
</div>
Second Part - This is the part I dont know how to do
<div class="col-md-6">
<center>
<img src="https://minotar.net/avatar/TOPVOTERHERE">
</center>
</div>

Answer 1

The simplest way to do it would be to create a variable to store the top voter data before you start your while loop. Then, in the while loop, you can set this variable if you're in the first iteration.
First part:
$result = mysql_query("SELECT * FROM GALTotals ORDER BY votes DESC");
$count = 0;
$topVoterIGN = NULL;
while ($row = mysql_fetch_assoc($result)) {
$count++;
if($count>10){
}else{
echo "<tr>";
echo "<td>".$row['IGN']."</td>";
echo "<td>".$row['votes']."</td>";
echo "</tr>";
if(is_null($topVoterIGN)) {
$topVoterIGN = $row['IGN'];
}
}
}
Second Part:
<div class="col-md-6">
<center>
<img src="https://minotar.net/avatar/<?php echo $topVoterIGN; ?>">
</center>
</div>
Alternatively, you can store the top 10 voters in an array and then retrieve the first item later.