displaying data for a single column from Mysql database in the list box in php form

<?php

$data = @mysql_query("select * from names");
echo "<p>Select a Name: n";
echo "<Select Name="ID">n";
while ($row = mysql_fetch_assoc($data))
{
$ID = $row['ID'];
$year = $row['name'];
echo "<option value=$ID>$namen";
}
echo "</select>n";
echo "</p>n";
?>

YOUR CODE CONTAIN ERROR..
<?php

$data = @mysql_query("select * from names");
echo "<p>Select a Name: n";
echo "<Select Name="ID">n"; 
while ($row = mysql_fetch_assoc($data))
{
$ID = $row['ID'];
$year = $row['name'];
echo "<option value=$ID>$namen";  // close option tag
}
echo "</select>n";
echo "</p>n";
?>


you should write following type code..

<?php       
include('include/functions.php');
$Select = "SELECT * FROM category_master";
$Q = mysql_query($Select);
?>
<form name="myForm" method="post">
    <select name="select"> 
<?php  while($row = mysql_fetch_assoc($Q))
        {

        echo "<option value=''> $row[_cat_name] </option>";

        }
?>  
        </select>
</form>

i am not getting the above code plz elaborate and explain

 <?php      
    include('include/functions.php');  // Include DB Connection File.
        $table = "categorymaster"; // Table Name
        $field1 = "catname"; // Field Nmae
        // Change the Above Table Name and Field Name.

    echo "<form name='myForm' method='post'>"; // Start Form Tag for Posting Element.
    echo "<select>"; // Select Tag

        $Select = "SELECT * FROM $table"; // Select Query.
        $Q = mysql _ query($Select);  // Run Query.

        while($row = mysql_ fetch_ assoc($Q)) // Fetch Array in $row
        {
                echo "<option>"; // Start Option Tag.
                echo "$row[$field1]"; // Fetch Value From DB.
                echo "</option>"; // End Option Tab.
        } 
    echo "</select>"; // End Option Tab.
    echo "</form>"; // End Form Tab.
    ?>