Drop Down Database Form

NCC1701

Having trouble creating functions to display a list of values in a drop down menu, all i want to display is the menu option, this will evetually be coded so that when the user selects the menu name, the menu is displyed in a text box and will be able to edit it.

<?php 
function get_menus()
{
  //query database
  $conn = db_connect();
  $query = "select menu_name
  			from menu
			where menu_id = $menu_id";
  $result = @mysql_query($query);
  if (!$result)
  	return false;
  $num_menus = @mysql_num_rows($result);
  if ($num_menus == 0)
    return false;
  $result = mysql_result($result, 0, 'menu_name');
  return $result;
}



function display_menu_form()
{
?>
  <form method="post" action="">
  <table align="center" border="0">
  <tr>
    <td><select name="menu_name">
	<?php 
	  //selects menu list from menu database
	  $menu_array=get_menus();
	  foreach ($menu_array as $menu)
	  {
	    echo '<option value="';
		echo $menu['menu_name'];
		echo '"';
	  }  

	?>
	</select></td>
  </tr>
  </table>
  </form>
<?php 
}
?>

NCC1701

i think i was trying to be complecated above, can anyone provide a very simple script to input database items into a drop down menu?

i already have a function that connects to a database called db_connect()

goldbug

you need a mysql_fetch_array() or mysql_fetch_row() in there somewhere.

Example from the manual:

<?php
    mysql_connect("localhost", "mysql_user", "mysql_password") or
        die("Could not connect: " . mysql_error());
    mysql_select_db("mydb");

$result = mysql_query("SELECT id, name FROM mytable");

while ($row = mysql_fetch_array($result)) {
    printf ("ID: %s  Name: %s", $row[0], $row[1]);  
}

mysql_free_result($result);
?>

devinemke

you need to pass $menu_id to get_menus() as an argument or declare it global within the function

function get_menus ($menu_id) 
{ 
  $conn = db_connect (); 
  $query = "SELECT menu_name FROM menu WHERE menu_id = $menu_id";
	$result = @mysql_query ($query);
	if ($result)
	{
	  if (mysql_num_rows ($result))
		{
			while ($row = mysql_fetch_assoc ($result)) {$array[] = $row['menu_name'];}
			return $array;
		}
		else {return false;}	  

	} 
	else {return false;}
} 

function display_menu_form ($menu_id)
{  

	echo '
	<form method="post" action="">
	<table align="center" border="0">
	<tr> 
	<td>
	<select name="menu_name">
	';

$menu_array = get_menus ($menu_id);
if ($menu_array)
{
	foreach ($menu_array as $menu) {echo '<option>' . $menu . "\n";}
}

echo ' 
</select>
</td> 
 </tr> 
 </table> 
 </form>
';
}

NCC1701

Thanks guys, its working now, really appreciate the help.

All i need now is that when the option in the menu is selected the appropriate menu text is displayed in a textarea to be edited then updated, someone mentioned that i HAD to use javascript 'onchange' but i dont wanna use javasript, there must be some way to do it using php, any ideas?

devinemke

if you want the menu text to appear instantly when the option is selected then must use javascript. to use PHP, you must submit the form.

NCC1701

right ive got that portion working now, now instead of adding javascript for when the option is chosen i need to sumbit the option and display another form so the menu can be edited, can anyone tell me how i would pass this chosen option on to another form?

thanks
NCC1701