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ffwpautzfeld
Funnilein


Joined: 30 Jan 2008
Posts: 1

PostPosted: 19.02.08 12:53    Post subject: Probleme mit Login Formular Reply with quote
Hallo ,
wenn ich mein Login Formular starte brint er immer SQL Fehler:

Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'user'@'192.168.101.2' (using password: YES) in /usr/export/www/hosting/ffwpautzfeld/login-formular-mit-php-und-mysql.php on line 11

Warning: mysql_select_db(): supplied argument is not a valid MySQL-Link resource in /usr/export/www/hosting/ffwpautzfeld/login-formular-mit-php-und-mysql.php on line 12

usw.
unter der WEB http://ffwpautzfeld.ff.funpic.org/

Welche Fehler mache ich, das Formular schaut so aus:


<?php

$DatabaseHost = "localhost";
$DatabaseUser = "user";
$DatabasePassword = "password";
$Database = "mydatabase";
$TableLogin = "Login";

if($_REQUEST['Send'])
{
$DatabasePointer = mysql_connect($DatabaseHost, $DatabaseUser, $DatabasePassword);
mysql_select_db($Database, $DatabasePointer);
if(mysql_num_rows(mysql_query("SELECT User FROM $TableLogin WHERE User='".stripslashes($_REQUEST['User'])."' AND Passwort='".stripslashes($_REQUEST['Passwort'])."'", $DatabasePointer))==1)
{
mysql_query("UPDATE $TableLogin SET LetzterLogin='".date("Y-m-d H:i:s")."' WHERE User='".stripslashes($_REQUEST['User'])."' AND Passwort='".stripslashes($_REQUEST['Passwort'])."'", $DatabasePointer);

echo"Sie sind eingeloggt!";

//hier der geschützte Inhalt

}
else
{
echo"Der User und/oder das Passwort wurden nicht gefunden!";
}
}
else
{
?>
<form action="<?php echo$_SERVER['PHP_SELF']; ?>" method="post">
User <input maxlength="50" name="User" type="text"><br>
Passwort <input maxlength="50" name="Passwort" size="10" type="password"><br>
<input name="Send" type="submit" value="Login">
</form>
<?php
}

?>
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crushmaster
Fun-Schreck


Joined: 26 Jul 2007
Posts: 171

PostPosted: 21.02.08 11:14    Post subject: Reply with quote
dir is aber schon klar, dass du deine mysql zugangsdaten hier eintragen musst?
Quote:
$DatabaseHost = "localhost";
$DatabaseUser = "user";
$DatabasePassword = "password";
$Database = "mydatabase";
$TableLogin = "Login";

ausserdem muss die datenbanktabelle existieren, in deinem fall "Login".
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