Hi,
Actually you have an undefined variable called $row which is in this
context ($row->Vorname) is an object...but $row has nor content
neither type.
I think you'd better use arrays...in this way:
(according to the PHP manual mysql_db_query() is decrepated)
Of course some error handling may c
'';
echo "";
echo "";
echo " Name";
echo '';
echo "";
...
?>
bastien
From: Ruprecht Helms <[EMAIL PROTECTED]>
Reply-To: [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: [PHP-DB] Scriptproblem
Date: Tue, 24 Jan 2006 16:12:53
Hi,
how can I display the stored data of an user as value in a formfield.
Actualy the formfield "vorname" shows no entry and the formfield "name"
shows .$row->Name.
What is the correct syntax in this case.
Regards,
Ruprecht
,);
$result=mysql_db_query("salzert","SELECT * FROM Benutzer WHER
