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'], $result['Name'])
?
Of course some error handling may come handy (was connect to the
database server / database selection succesful, was my query resulted
more than 0 rows or more than 1 rows or it's only produces an error
message). Check out the return values of the above functions and check
them in your script.
Hope it helped, bye:
Balazs
2006/1/24, Ruprecht Helms [EMAIL PROTECTED]:
Hi,
how can I display the stored data of an user as value in a formfield.
Actualy the formfield vorname shows no entry and the formfield name
shows .$row-Name.
What is the correct syntax in this case.
Regards,
Ruprecht
?
mysql_connect(localhost,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;
echotdVorname/td;
echo 'tdinput type=text name=vorname value=';
echo $row-Vorname;
echo '/td';
echo /tr;
echo tr;
echo tdName/td;
echo 'tdinput type=text name=name value=.$row-Name./td';
echo /tr;
...
?
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