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)
<?php
$link = mysql_connect(_host_, _user_, _pw_);
$db = mysql_select_db(_db_name_, $link);
$query = mysql_query("SELECT Vorname, Name FROM Benutzer WHERE ID = "
. $id, $link);
$result = mysql_fetch_assoc($query);
//and than you will get the desired values as an associative array
($result['Vorname'], $result['Name'])
?>
Of course some error handling may come handy (was connect to the
database server / database selection succesful, was my query resulted
more than 0 rows or more than 1 rows or it's only produces an error
message). Check out the return values of the above functions and check
them in your script.
Hope it helped, bye:
Balazs
2006/1/24, Ruprecht Helms <rhelms@xxxxxxxxxx>:
> Hi,
>
> how can I display the stored data of an user as value in a formfield.
> Actualy the formfield "vorname" shows no entry and the formfield "name"
> shows .$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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