You seem to be mixing OO and procedural coding styles...try

<?
mysql_connect("localhost",<user>,<password>);
$result=mysql_db_query("salzert","SELECT * FROM Benutzer WHERE ID=$id");

$row = mysql_fetch_array($result); //get the data into an array

echo '<input type="text" name=id value="';
echo $id;
echo '">';
echo "<tr>";
echo " <td>Vorname</td>";
echo '<td><input type="text" name="vorname" value="';
echo $row['Vorname'];
echo '"></td>';
echo "</tr>";
echo "<tr>";
echo " <td>Name</td>";
echo '<td><input type="text" name="name" value=".$row['Name']."></td>';
echo "</tr>";
...
?>



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 +0100

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

 <?
   mysql_connect("localhost",<user>,<password>);
  $result=mysql_db_query("salzert","SELECT * FROM Benutzer WHERE ID=$id");
   echo '<input type="text" name=id value="';
  echo $id;
echo '">';
  echo "<tr>";
  echo "   <td>Vorname</td>";
        echo '<td><input type="text" name="vorname" value="';
        echo $row->Vorname;
        echo '"></td>';
echo "</tr>";
echo "<tr>";
echo "     <td>Name</td>";
        echo '<td><input type="text" name="name" value=".$row->Name."></td>';
echo "</tr>";
...
?>

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