Use Of Undefined Constant in script

2019-03-06 18:44发布

问题:

I've searched the site and saw fixes for the issue in which the user should have put single quotes around the variable but I still a bit confused.

Errors: (all refer to line 28) Notice: Use of undefined constant log_id - assumed 'log_id' Notice: Use of undefined constant log_username - assumed 'log_username' Notice: Use of undefined constant log_password - assumed 'log_password'

Script:

     <?php
      session_start();

include ("includes/connection.php");

$user_ok = false;
$log_id = "";
$log_username = "";
$log_password ="";


function evalLoggedUser($con,$id,$u,$p){
    $sql = "SELECT ip FROM users WHERE id='$id' AND username='$u' AND password='$p' AND activated='1' LIMIT 1";
    $query = mysqli_query($con, $sql);
    $numrows = mysqli_num_rows($query);
    if($numrows > 0){
        return true;
    }
}

if(isset($_SESSION["userid"]) && isset($_SESSION["username"]) && isset($_SESSION["password"])) {
    $log_id = preg_replace('#[^0-9]#', '',$_SESSION['userid']);
    $log_username = preg_replace('#[^a-z0-9]#i','', $_SESSION ['username']);
    $log_password = preg_replace('#[^a-z0-9]#i','', $_SESSION ['password']);

    //VERIFY USER

    $user_ok = evalLoggedUser($con,log_id,log_username,log_password);

} else if(isset($_COOKIE["id"]) && isset($_COOKIE["user"]) &&  isset($_COOKIE["pass"])) {
    $_SESSION['userid'] =  preg_replace('#[^0-9]#', '',$_COOKIE['userid']);
    $_SESSION ['username'] =preg_replace('#[^a-z0-9]#i','', $_COOKIE ['username']);
    $_SESSION['password'] =  preg_replace('#[^a-z0-9]#i','', $_COOKIE ['username']);

    //CREATE LOCAL SHORT VARIABLES

    $log_id = $_SESSION['userid'];
    $log_username = $_SESSION['username'];
    $log_password = $_SESSION['password'];

    if($user_ok == true){}
    //UPDATE LAST LOGIN TIME
    $sql = "UPDATE users SET lastlogin=now() WHERE id='$log_id' LIMIT 1";
    $query = mysqli_query($con,$sql);
}


?>

回答1:

Change this line, you forgot the $ infront of the variables:

$user_ok = evalLoggedUser($con,$log_id,$log_username,$log_password);


回答2:

Your forgot to put dollar-signs.

Replace this:

$user_ok = evalLoggedUser($con,log_id,log_username,log_password);

with this:

$user_ok = evalLoggedUser($con,$log_id,$log_username,$log_password);


回答3:

$user_ok = evalLoggedUser($con,log_id,log_username,log_password);

change to:

$user_ok = evalLoggedUser($con,$log_id,$log_username,$log_password);


标签: php mysqli