Display only one entry from the database with PHP - php

Question

The following code retrieves and displays the correct data from the database however it gets all the data. I need a way to assign each value it retrieves from the database to a PHP variable. For example, if it gets "Joe", "Henry", and "Robert" from the database, I'd like one variable for each of those and right now it returns and array with all the values.
<?php
dbCon();
$query1 = mysql_query("SELECT * FROM hosts WHERE name!=''");
while($row1 = mysql_fetch_assoc($query1)) {
$res = $row1['name'] . '<br />';
echo $res;
}
?>

Answer 1

<?php
function echoName($id) {
dbCon();
$query1 = mysql_query("SELECT * FROM hosts WHERE id='$id'");
while($row1 = mysql_fetch_assoc($query1)) {
$res = $row1['name'] . '<br />';
echo $res;
}
}
?>
<div id="joe_div">
<?
echoName("1");
?>
</div>
.....
<div id="henry_div">
<?
echoName("2");
?>
</div>

Answer 2

Declare variable variables as such:
while($row1 = mysql_fetch_assoc($query1)) {
$$row1['name'] = $row1['name']
}
echo $Joe; //returns Joe
Though I don't know why you would ever need that

Answer 3

if it gets "Joe", "Henry", and "Robert" from the database, I'd like one variable for each of those
So try this:
<?php
dbCon();
$query1 = mysql_query("SELECT * FROM hosts WHERE name!=''");
while($row1 = mysql_fetch_assoc($query1)) {
$$row1['name'] = $row1['name'];
}
echo $Joe;
echo $Henry;
echo $Robert;
?>
P.S: I don't know why you want to do this, but i am sure you have a better approach to solve your problem.