为什么我不能通过PHP中的$ _POST存储选定的值

Question

I create a dropdown list to select values:

 <form action="." method="post" id="aligned">
    <input type="hidden" name="action" value="update_customer">
    <input type="hidden" name="customer_id" 
           value="<?php echo htmlspecialchars($customer['customerID']); ?>"> 


    <label>Country:</label>
    <select name="selected">
     <?php  

        $dbhost = 'localhost';
        $dbuser = 'root';
        $dbpass ='';
        $db = 'tech_support';

        $conn = mysql_connect($dbhost, $dbuser, $dbpass);
        if(!$conn)
            die('Could not connect: '. mysql_error());

        mysql_select_db($db);
        $selected= mysql_query("select * from countries where countryCode = '" .$customer['countryCode']. "'");
        $sql = mysql_query("select * from countries order by countryName");

        if($selectedrow = mysql_fetch_array($selected)){
             echo "<option selected value=\"VALUE\">" . $selectedrow['countryName']."</option>";
        }
         //echo "<select>";

        while ($row = mysql_fetch_array($sql)) {
        echo "<option value =\"VALUE\">". $row['countryName']."</option>";
        }
        //echo "</select>";
       ?> 
    </select><br>

In another php file as below, I was trying to store the selected value for Country, but nothing is stored, why?

<?php $country_name = $_POST["selected"];
echo $country_name;//it always print out 'VALUE'-I guess it means null.
?>

Answer 1

You should use

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

And I want to add something, MySQL is deprecated in PHP5 and removed from PHP7. what is your PHP version. If you use PHP 5 or later use mysqli.