pes96

My apologies, your sql syntax is correct :-\ First, I thought that the line 28 is just before the query I gave you above. Secondly, you can make a simple debugging test to check, what actually values you will send to database. <?php $username = 'jazzman'; $password = 'password'; $id = 1; $sql1 = "SELECT * FROM user_accounts WHERE username='".$username."' AND password='".$password."' AND user_id='".$id."'"; $sql2 = 'SELECT * FROM user_accounts WHERE username ='.$username.' AND password='.$password.' AND user_id='.$id; // integer id $sql3 = "SELECT * FROM user_accounts WHERE username='$username' AND password='$password' AND user_id='$id'"; $sql4 = 'SELECT * FROM user_accounts WHERE username=$username AND password=$password'; // invalid query printf('%s<br />%s<br />%s <br />%s', $sql1, $sql2, $sql3, $sql4); So and now? Thanks for your patience

So someone can guide me to what i have to do?

I think, you need to start learning php from beginning - step by step . But you can only help me correcting this code, (saying what i have to do) after i will learn php, but now i need this code working fine. Thanks for all

My apologies, your sql syntax is correct :-\ First, I thought that the line 28 is just before the query I gave you above. Secondly, you can make a simple debugging test to check, what actually values you will send to database. <?php $username = 'jazzman'; $password = 'password'; $id = 1; $sql1 = "SELECT * FROM user_accounts WHERE username='".$username."' AND password='".$password."' AND user_id='".$id."'"; $sql2 = 'SELECT * FROM user_accounts WHERE username ='.$username.' AND password='.$password.' AND user_id='.$id; // integer id $sql3 = "SELECT * FROM user_accounts WHERE username='$username' AND password='$password' AND user_id='$id'"; $sql4 = 'SELECT * FROM user_accounts WHERE username=$username AND password=$password'; // invalid query printf('%s<br />%s<br />%s <br />%s', $sql1, $sql2, $sql3, $sql4); So i have to put this code where?

Now appears: "Select db failed: Access denied for user 'a4199910_login'@'10.1.1.43' to database 'Login'"

What is your current code? And is there any chance you are using a cheap/free web host where changes you make to your files don't take affect immediately due to disk caching? Now i have done and appears the message: "No database selected"

So you can explain me where is the error in my code? Thanks for all

I do this, but appears the same error.

And i have to put this on line 28? Thanks

PHP Error Message Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource /home/a4199910/public_html/login2.php on line 28 http://teste-vieirasousa.tk/login2.php THE CODE: (the values of database, password, username, server that i use here are an example) <?php $link = mysql_connect("localhost", "username", "password"); mysql_select_db("database_name", $link); function login($username, $password) { $username = addslashes($username); $password = md5($password); $query = mysql_query("SELECT * FROM user_accounts WHERE username='$username' AND password='$password'"); if(mysql_num_rows($query) == 1) { $info = mysql_fetch_array($query); $userid = $info[userid]; $sessionid = md5($userid . time()); $time = time(); @setcookie ('test_account', $sessionid, $time+3600, '/', ''); mysql_query("DELETE FROM user_sessions WHERE userid='$userid'"); mysql_query("INSERT INTO user_sessions (sessionid,userid,timestamp) VALUES('$sessionid','$userid','$time')"); return $userid; } else { return 0; } } function status() { $sessionid = $_COOKIE[test_account]; $oldtime = time() - 3600; $query = mysql_query("SELECT * FROM user_sessions WHERE sessionid='$sessionid' AND timestamp>$oldtime"); if(mysql_num_rows($query) == 1) { $info = mysql_fetch_array($query); return $info[userid]; } return 0; } function logout() { $sessionid = $_COOKIE[test_account]; @setcookie ("test_account",'', time()-99999, '/', ''); mysql_query("DELETE FROM user_sessions WHERE sessionid='$sessionid'"); } if($_POST[username] !='' || $_POST[password] != '') { $login_status = login($_POST[username], $_POST[password]); } else if($_GET[logout]) { logout(); } $userid = status(); if($userid > 0) { echo "Welcome to our site, user #$userid (<a href='?logout'>Click here to logout</a>)"; } else { if($login_status != '' && $login_status == 0) { echo "Invalid username/password combo.<br>"; } ?> <form action="sample.php" method="POST"> <input type=text name=username> <input type=password name=password> <input type=submit value="Log In"> </form> <?php } ?> LINE 28: if(mysql_num_rows($query) == 1) { Anyone can help me to resolve this error? Thank you