php mysql dropdown value select - php

Question

Currently the following code displays the filters for the individual publishers ie if I select DC I will only see the DC titles.
The issue is that from the current code I cannot get it to display all publishers from the dropdown menu.
<div id="main">
<form method="post" action="">
<div id="search_query" >
<select name="make" size="0">
<option value="all">All Publishers</option>
<option value="DC">DC</option>
<option value="Marvel">Marvel</option>
<option value="Image">Image</option>
</select>
<input type="submit" name="submit" value="submit">
</div>
</form>
<div id="main_container">
<?php
$db_con = mysql_connect('127.0.0.1','root','root');
if (!$db_con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db('work', $db_con);
if(isset($_POST['submit']))
{
$make = mysql_real_escape_string($_POST['make']);
$sql = sprintf("SELECT * FROM products WHERE publisher= '$make' ");
$result = mysql_query($sql);
echo "<table width= 970 border=1>
<tr>
<th width='120' scope='col'>Title</th>
<th width='170' scope='col'>Publisher</th>
<th width='185' scope='col'>Price</th>
<th width='126' scope='col'>Desc</th>
</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>".$row['title']."</td>";
echo "<td>". $row['publisher'] . "</td>";
echo "<td>". $row['price'] ."</td>";
echo "<td>". $row['desc'] ."</td>";
echo "</tr>";
}
echo "</table>";
}
else
{
$sql = sprintf("SELECT * FROM products");
$result = mysql_query($sql);
echo "<table width= 970 border=1>
<tr>
<th width='120' scope='col'>Title</th>
<th width='170' scope='col'>Publisher</th>
<th width='185' scope='col'>Price</th>
<th width='126' scope='col'>Desc</th>
</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>".$row['title']. "</td>";
echo "<td>". $row['publisher'] . "</td>";
echo "<td>". $row['price'] ."</td>";
echo "<td>". $row['desc'] ."</td>";
echo "</tr>";
}
echo "</table>";
}
mysql_close($db_con);
?>

Answer 1

In your query, it's searching for publisher = $make where $make = "all".
Your database table probably doesn't have any record with publisher = "make".
I'd suggest you to add an if else condition here:
if (isset($_POST['submit'])) {
$make = mysql_real_escape_string($_POST['make']);
if ($make != "all") {
$sql = sprintf("SELECT * FROM products WHERE publisher= '$make' ");
} else {
$sql = sprintf("SELECT * FROM products");
}
/* Other code */
}