Question

What i’m trying to do is display a drop down with all field names from mysql database, once the user picks one and submits the form i want to display a second dropdown filled with all the rows from the submitted field name, this is my code so far:

$result = mysql_query("select * from `parts`") or die(mysql_error());

echo "<form action='".$_SERVER['PHP_SELF']."' method='post'>";
echo "<select name='field_names'>";

$i = 0;
while ($i < mysql_num_fields($result)) {
  $fieldname = mysql_field_name($result, $i);
  echo '<option value="'.$fieldname.'">'.$fieldname.'</option>';
  $i++;
}

echo "</select>";
echo "<input type='submit' value='submit'></input>";
echo "</form>";

if($_POST) {
  $fields = $_POST['field_names'];
  $result1 = mysql_query("select '".$fields."' from `parts`") or die(mysql_error());
  echo '<select name="fields">';

  while ($row = mysql_fetch_array($result1)) {
    echo "<option value=".$row[$fields].">".$row[$fields]."</option>";
  }
  echo '</select>';
}

Can anyone spot where i’m going wrong, thanks

Answer 1

As it stands now, the second set of selects will be issued OUTSIDE of your </form> tag, so will never get submitted with the rest of the form. At best, you should move the form closing tag to below the POST handler.