Question

I need a script that loads data form mysql and show on a drop down list. From there, I need to pass the selected data to another page.
I have done the first step. I am now able to load data from mysql table and to show them on a drop down menu. The exact is given bellow.

<?php
include("config.php");
$result= mysql_query("SELECT folder_name FROM folders");

echo '<select name="directory">'; // Open your drop down box

while ($row = mysql_fetch_array($result)) {
   //echo "<option>" . $row['folder_name'] . "</option>";
      echo '<option value="'.$row['folder_name'].'">'.$row['folder_name'].'</option>';
}
echo '</select>';// Close your drop down box
?>

Now I need help to pass the selected data to another page. Any suggestion please?

Answer 1

Let me consider the form is posted to page2.php from page1.php

page1.php

<form method="post" action="page2.php">
        //your dropdown code here
        <?php
              include("config.php");
              $result= mysql_query("SELECT folder_name FROM folders");

              $str = '';   
              $str .= '<select name="directory">'; // Open your drop down box

              while ($row = mysql_fetch_array($result)) {
              $str .= '<option    value="'.$row['folder_name'].'">'.$row['folder_name'].'</option>';
              }
              $str .= '</select>';// Close your drop down box

              echo $str;
         ?>

   <input type="submit" value="submit" />
</form>

in page2.php you can access the dropdown selected value as

$selVal = '';
if(isset($_POST['directory']))
{
    $selVal = $_POST['directory'];
}