PHP Login Script result resource

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Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result

When typing in a username and password im getting the error

mysql_num_rows(): supplied argument is not a valid MySQL result resource

IT IS SAYING I HAVE NO DATABASE SELECTED ...

I was wondering what this error means. Could it be a problem connecting to the database or is it a syntax problem. here is the code for the username and password

$query = mysql_query("SELECT * FROM Users WHERE username = '$user'");
                $numrows = mysql_num_rows($query);
                if($numrows == 1){
                    $row = mysql_fetch_assoc($query);
                    $dbid = $row['id'];
                    $dbuser = $row['username'];
                    $dbpass = $row['password'];
                    $dbactive = $row['active'];

                    if($password == $dbpass){
                        if($dbactive == 1){

                            //set session info
                            $_SESSION['userid'] = $dbid;
                            $_SESSION['username'] = $dbuser;

                            echo "You have been logged in as <b>$dbuser</b> Click here to go to member page.";

                        }else
                            echo "You must activate your account to login.";

                    }else 
                        echo "You did not enter the correct password.";


                }else 
                    echo "The username you entered was not found.";
            }


        }else
            echo "You must enter your username.";

    }

The problem is in the $query line. the first line

`var_dump("SELECT * FROM Users WHERE username = '$user'");` + `mysql_error();` — zerkms, Apr 22 '12 at 20:54
Your query has failed. do `$result = mysql_query(...) or die(mysql_error())` to find out why. — Marc B, Apr 22 '12 at 21:00

Lawrence Cherone · Answer 1 · 2012-04-22T21:31:23.707

The most likely cause is that your not connected or not selecting the correct database, you should use mysql_error() to track this down. But I also see some logic confusion, here is an example of a super simple login, checks should be made on the database query, not by PHP (plus there is no need to store the password anywhere other then in the database). You should also hash user account passwords & escape user input, hope it helps..

<?php 
session_start();
$db = mysql_connect('host','uesr','pass') or die(mysql_error());
mysql_select_db('database') or die(mysql_error());

if($_SERVER['REQUEST_METHOD'] == 'POST' && isset($_POST['username']) && isset($_POST['password'])){

    $username = mysql_real_escape_string($_POST['username']);
    $password = mysql_real_escape_string($_POST['password']);

    $query = mysql_query("SELECT `username`,`active`
                          FROM Users 
                          WHERE username ='$user' && password='$password' && active=1
                          LIMIT 1");

    if(mysql_num_rows($query) == 1){
        //Log in user found
        $row = mysql_fetch_assoc($query);

        $_SESSION['username']  = $row['username'];
        $_SESSION['active']    = $row['active'];
        $_SESSION['logged_in'] =true;
        header('Location: ./members.php');
        die;
    }else{
        //failed login
        $_SESSION['logged_in']=false;
        $_SESSION['log_in_error']='Fail!! Account not active or incorrect login information!';
        header('Location: ./login.php');
        die;
    }
}
?>

what is the name of your database? NOT table, perhaps rename the database to something other then a table within it. then you wont get confused. — Lawrence Cherone, Apr 22 '12 at 21:34
Also are you using `mysql_select_db('database') or die(mysql_error()); ` I notice @Terry Seidler answer does not include this. — Lawrence Cherone, Apr 22 '12 at 21:37

score 0 · Answer 2 · edited Apr 22 '12 at 21:44

0

Like zerkms suggested, try printing the mysql_error() to screen. Something like:

$query = mysql_query("your_query here");
if(!$query){
    echo 'Error! : ' . mysql_error();
}

http://www.php.net/manual/en/function.mysql-error.php

Before you do all of this, make sure you are connected to a database. For example: http://php.net/manual/en/function.mysql-connect.php

$link = mysql_connect('localhost', 'database_user', 'database_password');
if(!$link){
    die('Could not connect: ' . mysql_error());
}

//select your db
mysql_select_db("YOUR_DB_NAME",$link);


$result = mysql_query("SELECT * FROM Users WHERE username = '$user'");
if(!$result){
    echo 'Error! ' . mysql_error();
}

edited Apr 22 '12 at 21:44

esertbas

476
3
7

answered Apr 22 '12 at 20:59

Terry Seidler

2,043
15
28

I did what you said and the echo said that i have no database selected. – Waggoner_Keith Apr 22 '12 at 21:04
Have you connected to a database? http://www.php.net/manual/en/function.mysql-connect.php – Terry Seidler Apr 22 '12 at 21:06
Yea i have in another script it look like this mysql_connect("hostname","username","password"); mysql_select_db("Users"); – Waggoner_Keith Apr 22 '12 at 21:12
Then in the other script i write require(connect.php) – Waggoner_Keith Apr 22 '12 at 21:13
And the mysql_connect works? I have edited my first post with some more mysql_connect 'debugging'. – Terry Seidler Apr 22 '12 at 21:14
1

it the database called `Users`? dont confuse that with the table. – Lawrence Cherone Apr 22 '12 at 21:15
Good point by Lawrence ;) Just like with mysql_query and mysql_connect, you can also verify the result of mysql_select_db(). It returns TRUE on success and FALSE when it fails. – Terry Seidler Apr 22 '12 at 21:18
I did the test and i successfully connected to the database that is not the problem i guess any other ideas? – Waggoner_Keith Apr 22 '12 at 21:20
Check my answer to compare it with your code. ;) – Lawrence Cherone Apr 22 '12 at 21:21
Nevermind i changed it and still got the errors. that isn't the problem – Waggoner_Keith Apr 22 '12 at 21:21
I am still having this same problem does anyone have any ideas????? – Waggoner_Keith Apr 22 '12 at 21:31

PHP Login Script result resource

2 Answers2