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I am currently working on a log-in page which accesses a mysql database with a user and password table. Whenever I try to run my code I get the following error no matter what I comment:

Parse error: syntax error, unexpected '=', expecting ',' or ';' in /home/scs/licenta/an2/gr927/rpie1814/public_html/testphpathome/login.php on line 15

Here is the main page in HTML:

    <!DOCTYPE html>
<html>
<head>
    <title>
        Gaming site
    </title>

</head>
<body>

<h1>
    Log-In
</h1>

<form action="login.php" method="post">
    Username: <br>
    <input type="text" name="user"> <br>
    Password: <br>
    <input type="text" name="pass"><br>

    <input type="submit" value="Submit">
</form>

</body>
</html>

And here is the php code:

    <?php

$user = $_POST["user"];
$pass = $_POST["pass"];


global $conn = mysqli_connect("localhost", "rpie1814", "rpie1814", "rpie1814");

if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

$sql = "SELECT * FROM Users";
global $result = mysqli_query($conn, $sql);

if(mysqli_num_rows($result) > 1)
{
    header("Location: /game.php");
    exit;
}
else
{
    echo "Unknown user";
}

mysqli_close($conn);
exit;

?>

2 Answers2

1

You have to remove global from connection ($conn) and result ($result):-

$conn = mysqli_connect("localhost", "rpie1814", "rpie1814", "rpie1814");

AND

Replace:

$result = mysqli_query($connect, $sql);

To:-

$result = mysqli_query($conn, $sql); // actually your connection variable is $conn not $connect
0

Replace:

global $result = mysqli_query($connect, $sql);

by:

$result = mysqli_query($conn, $sql);
Kurt Van den Branden
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