How to check if user exists in SQL database

Question

I want to check if the user exists in the database when clicking 'log in'. If so, echo "user exists" and if not "user does not exist". How can i go about this? so far i have:

<?php
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "test";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
     die("Connection failed: " . $conn->connect_error);
} 

?>

<!DOCTYPE html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Test</title>
</head>

<body>
<form method="post" action="login.php">
Username: <input type="text"  value="" name="username" /><br /><br />
Password: <input type="password" value="" name="password" /><br /><br />
<input type="submit" name="submit" value="Log In"/>
</form>
</body>
</html>

Hint: SELECT COUNT(*) ...

zerkms
– zerkms

2014-12-03 21:18:00 +00:00
Commented Dec 3, 2014 at 21:18 — zerkms
– zerkms, Commented Dec 3, 2014 at 21:18

Oleg · Accepted Answer · 2014-12-03 21:57:08Z

2

Try this one

session_start();
if($_SERVER['REQUEST_METHOD'] === 'POST'){

    $servername = "localhost";
    $username = "root";
    $password = "root";
    $dbname = "test";

// Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    }

    $user = $conn->real_escape_string(htmlspecialchars(trim($_POST['username'])));
    $query = "SELECT `username` FROM `users` WHERE `username` = '$user'";

    $result = $conn->query($query);
    if($result->num_rows > 0) {
       $_SESSION['allowToWelcome'] = true;
       header('Location:welcome.php');
       die();
    }
    else $message = 'user does not exist';

}

?>

    <!DOCTYPE html>
    <html>
        <head>
            <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
            <title>Test</title>
        </head>

        <body>
            <form method="post" action="login.php" enctype="multipart/form-data">
                <?php if(isset($message)) : ?>
                    <div class="error"><?php echo $message; ?></div>
                <?php endif; ?>

                Username: <input type="text"  value="" name="username" /><br /><br />
                Password: <input type="password" value="" name="password" /><br /><br />
                <input type="submit" name="submit" value="Log In"/>
            </form>
        </body>
    </html>

in welcome.php you can check like this

session_start();
if(!isset($_SESSION['allowToWelcome'])){
   header('Location:login.php');
   die();
}

edited Dec 3, 2014 at 21:57

answered Dec 3, 2014 at 21:26

Oleg

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4 Comments

HuskaV Over a year ago

I want to take the user to a 'welcome page' if they exist, if they do not, display 'user does not exist'. How do i link to another page? do i need to do this in the html, form action?

Oleg Over a year ago

Modified code above. Check line header('Location:welcome.php'); You can change welcome.php to any page in your project.

HuskaV Over a year ago

thanks again! for some reason, this still lets in-existing users to access the welcome.php for example..

Oleg Over a year ago

Modified again, last time. I think I've gave you full answer

knocked loose · Accepted Answer · 2014-12-03 21:28:39Z

1

On my login/register script I made, I used something along the lines of this..

// check existing username
$prep_stmt = "SELECT id FROM your-table-name WHERE username = ? LIMIT 1";
$stmt = $mysqli->prepare($prep_stmt);

if ($stmt) {
    $stmt->bind_param('s', $username);
    $stmt->execute();
    $stmt->store_result();

            if ($stmt->num_rows == 1) {
                    // A user with this username already exists
                    $error_msg .= '<p class="error">A user with this username already exists</p>';
                    $stmt->close();
            }
            $stmt->close();
    } else {
            $error_msg .= '<p class="error">Database error line 55</p>';
            $stmt->close();
    }

edited Dec 3, 2014 at 21:28

answered Dec 3, 2014 at 21:22

knocked loose

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3 Comments

DarkBee Over a year ago

Should add the mysqli prepare function and execute :)

zerkms Over a year ago

@DarkBee it might be a homework for OP actually :-) Even a query is already too much for a newbie

knocked loose Over a year ago

Sorry, I am just learning PHP / MySQL, and only know so much, figured I'd try to help someone out. Added my full function and if/else conditional statement.

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