Fatal error: Call to a member function query() on null

Question

I'm not sure what's going wrong here. I was just following a tutorial online and these errors popped up.

I'm getting the following errors

Error

Notice: Undefined variable: db in C:\xampp\htdocs\wisconsindairyfarmers\admin\login.php on line 7

Fatal error: Call to a member function query() on null in C:\xampp\htdocs\wisconsindairyfarmers\admin\login.php on line 7

Code

<?php
$db = new mysqli('127.0.0.1', 'root', '', 'wisconsindairyfarmers');
?>

<?php
require '../db/connect.php';
require '../functions/general.php';

    function user_exists($username){
        //$username = sanitize($username);
        $result = $db->query("SELECT COUNT(UserId) FROM users WHERE UserName = '$username'");
        if($result->num_rows){
        return (mysqli_result($query, 0) == 1) ? true : false;
    }}

if(empty($_POST) === false){

    $username = $_POST['username'];
    $password = $_POST['password'];

    if(empty($username) === true || empty($password) === true){ 
        echo 'You need to enter a username and password';
    }
    else if(user_exists($username) === false) {
        echo 'We can\'t find that username.';
    }
}

?>

Your `$db` variable is inside a function and thus out of scope from the code that defines it. Declare it global, or better, pass it as an argument to your function. See the PHP manual on [scope](http://php.net/manual/en/language.variables.scope.php) — , Jun 23 '15 at 02:22
I changed the variables, and the first error is fixed, but I'm still getting : Fatal error: Call to undefined function mysqli_result() — MPStimpson, Jun 23 '15 at 02:32
I think that is because you are using OOP on the first part and procedural for the the results. — Rasclatt, Jun 23 '15 at 02:38
I think that's because `mysqli_result()` is not a function. I'm really not clear what that piece of code is trying to do, but `mysqli_result()` doesn't exist in PHP. — , Jun 23 '15 at 02:41
It's suppose to check and see if a user exists in a table and if the user does exist it returns a 1. Is there another way to do it with mysqli? — MPStimpson, Jun 23 '15 at 02:43

score 24 · Accepted Answer · edited Nov 17 '22 at 22:11

24

First, you declared $db outside the function. If you want to use it inside the function, you should put this at the begining of your function code:

global $db;

And I guess, when you wrote:

if($result->num_rows){
        return (mysqli_result($query, 0) == 1) ? true : false;

what you really wanted was:

if ($result->num_rows==1) { return true; } else { return false; }

which is equivalent to:

return $result->num_rows == 1;

edited Nov 17 '22 at 22:11

William HUANG

3
4

answered Jun 23 '15 at 03:03

Juan Carlos Alvez Balbastro

535
5
9

I'm not getting any errors anymore, but I can't seem to get into the 'user_exists($username) === false' when I'm testing do you have any other wisdom to share with me today :) – MPStimpson Jun 23 '15 at 04:08
Yes, your SQL query will always return 1 row, with 1 field saying the amount of users equals to username. Change that to `"SELECT * FROM users WHERE UserName = '$username'"` and it will work – Juan Carlos Alvez Balbastro Jun 24 '15 at 17:41
I figured that out, but thank you very much! You've been very helpful! That's what I get for blindly following tutorials – MPStimpson Jun 26 '15 at 01:06

score 0 · Answer 2 · answered Aug 29 '16 at 22:59

put this line in parent construct : $this->load->database();

function  __construct() {
    parent::__construct();
    $this->load->library('lib_name');
    $model=array('model_name');
    $this->load->model($model);
    $this->load->database();
}

this way.. it should work..

Fatal error: Call to a member function query() on null

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