Re: [PHP-DB] Scriptproblem

[Balazs Hegedus]

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 come handy (was connect to the
database server / database selection succesful, was my query resulted
more than 0 rows or more than 1 rows or it's only produces an error
message). Check out the return values of the above functions and check
them in your script.

Hope it helped, bye:

Balazs

2006/1/24, Ruprecht Helms <[EMAIL PROTECTED]>:
> 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 .$row->Name.
>
> What is the correct syntax in this case.
>
> Regards,
> Ruprecht
>
>mysql_connect("localhost",,);
>$result=mysql_db_query("salzert","SELECT * FROM Benutzer WHERE ID=$id");
> echo '';
>echo "";
>echo "   Vorname";
> echo '';
> echo "";
> echo "";
> echo "  Name";
> echo '';
> echo "";
> ...
> ?>
>
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> PHP Database Mailing List (http://www.php.net/)
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>
>

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RE: [PHP-DB] Scriptproblem

[Bastien Koert]

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

,);
$result=mysql_db_query("salzert","SELECT * FROM Benutzer WHERE ID=$id");

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

echo '';
echo "";
echo " Vorname";
echo '';
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 +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

 ,);
  $result=mysql_db_query("salzert","SELECT * FROM Benutzer WHERE ID=$id");
   echo '';
  echo "";
  echo "   Vorname";
echo '';
echo "";
echo "";
echo " Name";
echo '';
echo "";
...
?>

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