display users selection

[obrienkev]

hi all,

I have created a select menu that displays results from a database.
The user selects an option from the menu and hits the Go button.
How do I display the users selection?(details in database)

Thanks.

My Select menu code::

echo '
		<form id="FormName" action="view_result2.php" method="post" name="FormName">';

	echo   '<select name="menu" size="1">';
	for ($i=0; $i<$num_results; $i++)
	{
		$row = mysql_fetch_array($result);
			echo '<option value="'.($i+1).'">'.$row['name'].'</option>';
	}
	echo '</select>';
	echo '<input type="submit" name="submit" value="Go">';
	echo '</form>';

Here is my code to display users selection::

mysql_select_db('test') or die(mysql_error());
		$query = "select name from answers where ".$menu." = name";
		$result = mysql_query($query);

	$num_results = mysql_num_rows($result);

	for ($i=0; $i<$num_results; $i++)
	{
	$row = mysql_fetch_array($result);
	echo 'Name selected: '.$row['name'];
	}

[saraadmin]

Do you mean that your selection code doesn't work?

If so, you might want to try $POST['menu'] rather than just $menu:

mysql_select_db('test') or die(mysql_error()); 
        $query = "select name from answers where name=\"".$_POST['menu']."\""; 
        $result = mysql_query($query); 

    $num_results = mysql_num_rows($result); 

    for ($i=0; $i<$num_results; $i++) 
    { 
    $row = mysql_fetch_array($result); 
    echo 'Name selected: '.$row['name']; 
    }