cty Posted December 18, 2006 Share Posted December 18, 2006 my problem is :whenever i log in ,it always shown 'You are not log in'.Never show "You are logged in as :xxx"For example:i do a testing,i input username:abc and password:1234 in "user" table.Next,i try to log in using abc and 1234.But,still shown 'You are not log in'can anyone tell me which part have bugs?or how i edit the code?I already use a phpeditor to check bugs,but no warning shown.Hope u able to help me.TQ ------------------------------------------------------------------------------//login.html<html><head><title>Login here</title></head><body><form method="POST" action="login.php"> Username <input type="text" name="username" size="20"></p> Password <p><input type="text" name="password" size="20"></p> <input type="submit" value="Submit" name="B1"> <input type="reset" value="Reset" name="B2"></p></form></body></html>----------------------------------------------------------------------//login.php<?phpsession_start();if(isset($_POST['username'])&&isset($_POST['password'])){$username=$_POST['username'];$password=$_POST['password'];$db=new mysqli('localhost','root','','kelly'); if(mysqli_connect_errno()){echo 'Connection to database failed:'.mysqli_connect_error();exit();}$query='select* from user '."where username='$username'" ."and password=sha1('$password')";$result=$db->query($query);if($result->num_rows >0){$_SESSION['valid_user']=$username;}$db->close();}?><html><body><h1>Home Page</h1><?phpif(isset($_SESSION['valid_user'])){ echo'You are logged in as :'.$_SESSION['valid_user'].'<br />';}else{ echo'You are not log in';}?>-----------------------------------------------------------------------//end Link to comment https://forums.phpfreaks.com/topic/31057-coding-bugsplz-help/ Share on other sites More sharing options...
trq Posted December 18, 2006 Share Posted December 18, 2006 When you [i]input[/i] your abc user into the database are you sure to sha1 the password? Link to comment https://forums.phpfreaks.com/topic/31057-coding-bugsplz-help/#findComment-143404 Share on other sites More sharing options...
redarrow Posted December 18, 2006 Share Posted December 18, 2006 [code]//login.html<html><head><title>Login here</title></head><body><form method="POST" action="login.php"> Username <input type="text" name="username" size="20"></p> Password <p><input type="text" name="password" size="20"></p> <input type="submit" value="Submit" name="B1"> <input type="reset" value="Reset" name="B2"></p></form></body></html>----------------------------------------------------------------------//login.php<?phpsession_start();if($_POST['B1']) {$username=$_POST['username'];$password=$_POST['password'];$db=new mysqli('localhost','root','','kelly'); if(mysqli_connect_errno()){echo 'Connection to database failed:'.mysqli_connect_error();exit();}$query='select* from user '."where username='$username'" ."and password=sha1('$password')";$result=$db->query($query);if($result->num_rows >0){$_SESSION['valid_user']=$username;}$db->close();}?><html><body><h1>Home Page</h1><?phpif(isset($_SESSION['valid_user'])){ echo'You are logged in as :'.$_SESSION['valid_user'].'';}else{ echo'You are not log in';}?>[/code] Link to comment https://forums.phpfreaks.com/topic/31057-coding-bugsplz-help/#findComment-143407 Share on other sites More sharing options...
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