On May 29, 2010, at 12:02 AM, Nathan Nobbe wrote:
On Fri, May 28, 2010 at 7:43 PM, Jason Pruim <lists@xxxxxxxxxxxxxxxxxxxx
> wrote:
Hey Everyone,
So I'm sitting here on a friday night trying to figure out how in
the world I'm going to fix an issue that should probably be simple
to me but is escaping me at the moment....
Take this authentication function:
<?PHP
function authentication($authUser, $authPass, $cfgtableAuth){
// Keep in mind, PASSWORD has meaning in MySQL
// Do your string sanitizing here
// (e.g. - $user = mysql_real_escape_string($_POST['user']);)
$authUser = mysql_real_escape_string($_POST['txtUser']);
$authPass = mysql_real_escape_string($_POST['txtPass']);
$md5pass = md5($authPass);
$loginQuery = "SELECT * FROM {$cfgtableAuth} WHERE
userLogin='".$authUser."' AND userPass='".$md5pass."' LIMIT 0,1;";
$loginResult = mysql_query($loginQuery) or die("Wrong
data supplied or database error" .mysql_error());
$row1 = mysql_fetch_assoc($loginResult);
if($row1['access'] == "5000000"){
foreach (array_keys($_SESSION) as $key)
unset($_SESSION[$key]);
die('account disabled');
}
if(is_array($row1)){
$_SESSION['userInfo'] = array( "userLogin" =>
$row1['userName'], "loggedin" => TRUE, "userName" =>
$row1['userName'], "userPermission" => $row1['userPermission']);
error_log("User has logged in: ".
$row1['userLogin']);
}else{
//$_SESSION['userInfo'] =array("loggedin" =>
FALSE);
die('authentication failed');
}
return TRUE;
}
?>
Here is how I am displaying the login form:
<?PHP
session_start();
$link = dbconnect($server, $username, $password, $database);
$page = $_GET['page'];
echo <<<CSS
<body>
<div class="contentwrapper">
CSS;
include("nav.php");
if ($_SESSION['userInfo']['loggedin'] == TRUE) {
MAIN PAGE DISPLAY HERE
}else{
//Display login info
echo <<<FORM
<div class="dark">
<form method="post">
<p>
You must login to proceed!<BR />
User Name: <input type="text" size="20"
name="txtUser"><BR />
Password: <input type="password" size="20"
name="txtPass"><BR />
<input type="submit" value="Login"><BR />
</p>
</form>
</div>
FORM;
if(isset($_POST['txtUser'])) {
$authUser = $_POST['txtUser'];
$authPass = $_POST['txtPass'];
$auth = authentication($authUser, $authPass, $cfgtableAuth);
}
}
?>
Now... the authentication actually works, and it logs me in
properly, but I have to click the login button twice.... Ideally I
should just do it once, so I'm wondering if anyone can spot my
grievous misstep here?
it looks to me like you need to move the authentication() call
if(isset($_POST['txtUser'])) {
$authUser = $_POST['txtUser'];
$authPass = $_POST['txtPass'];
$auth = authentication($authUser, $authPass, $cfgtableAuth);
}
above the check to see if the user has logged in, right after the
include("nav.php");
line. right now, when the user submits the form, your code is first
finding that the user isnt logged in, spitting out the 'please log
in' portion of the html then logging them in, so youre actually
already logged in when the form shows itself the second time!
Hey nathan,
You were close actually... :) If I moved just the $auth call it came
up and said that the auth failed... BUT if I moved that entire if
block to just below the include("nav.php"); line it works as it should!
Thanks for the pointer in the right direction! :)
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