Fout Bij Het Update

Door Jeroen op 17-11-2006 00:42

488 views

ok als ik doormiddel van mysql query me database wil update met het strax komende script veranderd hij der maar 1 inplaats van alle aangegeven dingen welke hij moet verandere?

iemand tips hoe ik het kan oplosse?

form.php


<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body>
<form action="update.php" method="post">
<?
	 			
	
			mysql_connect('localhost','root','');
			mysql_select_db('wie');
	
				$result = mysql_query("SELECT * FROM test WHERE naam = '1'");
				$row = mysql_fetch_assoc($result);
			
					if($result == 0)
						{
							echo 'db error<br>';
							echo mysql_error();
						}
					else
						{
							echo '<input type="text" name="lol" value="'.$row['lol'].'"><br>
								  <input type="text" name="test" value="'.$row['test'].'"<br>
								  <input type="text" name="about" value="'.$row['about'].'"<br>';
						}
		
?>
<br><input type="submit">
</form>
</body>
</html>

update.php


<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body>
<?
if(isset($_POST['lol']))
		{
			mysql_connect('localhost','root','');
			mysql_select_db('wie');
			
				$result = mysql_query("UPDATE test SET lol=".$_POST['lol'].",test=".$_POST['test'].",about=".$_POST['about']." WHERE naam='1'");
					if($result == 0)
						{
							echo 'there was a problem<br>';
							echo mysql_error();
						}
					else
						{
							echo 'succesfull';
						}
		}
		?>
</body>
</html>