subha rr Posted December 17, 2008 Share Posted December 17, 2008 Hi...Below the code<?phpinclude "pass.php";session_start();$user = isset($_POST['user']) ? $_POST['user'] : $_SESSION['user'];$pass = isset($_POST['pass']) ? md5($_POST['pass']) : $_SESSION['pass'];mysql_query("insert into login('', '$user', '$pass')");$link = mysql_connect($hostname, $username, $password);if($link){$db = mysql_select_db("time", $link);}$sql = "select * from login where username = '$user' and password = password('$pass')";$result2 = mysql_query($sql);if(!isset($pass) || !isset($user)){if($result2){?><form method=POST action=""><div align=center >Enter login details</div><br><table align = center><tr><td>Username</td><td><input type=text name=user value=""><tr><td>Password</td><td><input type=password name=pass value=""><tr><td><input type=submit value="Submit"></td></tr></table></form><?php}}else if($result2 != 'pass'){echo "<font>Authentication Failed</font>";}?>issue for this code for password is correct or incorrect, the result will be occured Authentication failed..What is the issue for this code?pls help..regardssubha Link to comment Share on other sites More sharing options...
Synook Posted December 17, 2008 Share Posted December 17, 2008 The $result2 variable is a MySQL resource - you need to use mysql_fetch_assoc() or similar to get the actual value. Link to comment Share on other sites More sharing options...
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