使用$ _POST将所选项目值发送到服务器

i have a dropdown and the items of the dropdown Filling dynamic, i looking for a way to send selected value to server using php not ajax i have a <form> and submit button, have to access the selected value on server, this is my code:

<form action="singer.php" method="post">
  <table>
    <tr>
      <td>سبک</td>
      <td><select name="genre_list">
          <?php
  include('../db_inc.php');
$sql="select * from genre";
$result = mysql_query($sql) or die(mysql_error());
while($row = mysql_fetch_object($result)){
    $option =$row->genre_name;
    $value =$row->genre_id;
    echo '<option value='.$value.'>'.$option.'</option>';

    }
?>

        </select></td>

  </table>
</form>

i'm googled this issue but i can't find the solution,

If you submit to the current file, you will see the selected form by using $_POST['genre_list']

If you want to send the user selected value, you should check where does $_POST['genre_list'] appears at the while-loop and then add to his <option> the attribute selected="selected" so it should look like

$isSelected = ($_POST['genre_list'] == $row['COLUMN-NAME'] ? "selected='selected' " : "");
echo '<option $isSelected value='.$value.'>'.$option.'</option>';

Do not forget to avoid SQL-Injection when getting values from user client.

have it output the variable for the value surrounded with quotes "

echo '<option value="'.$value.'">'.$option.'</option>';

then use: $_POST['genre_list'] to show the selected genre_id

In singer.php, $_POST['genre_list'] should contain the value from the SELECT.