Почему не работает mysqli_fetch_array?

Question

[Stoble]

Ошибка:Warning
: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in
D:\OpenServer\domains\Site\products.php
on line
431

Строка 431 - while($row = mysqli_fetch_array($result)) { ?>

<?php
include_once "php/connect.php";


?>

<?php
                        
                        $query = "SELECT * FROM products";
                        $results = mysqli_query($connect, $query);


                        while($row = mysqli_fetch_array($result)) { ?>

                        
                            <div class="product__block">
                                    <div class="product__block--img">
                                        <img src="images/<=? $row ['image']?>" alt="">
                                    </div>
    
                                    <div class="product__block--info">
                                        <div class="product__block--info--info">
                                            <div class="product__block--title"><?= $row['name']; ?></div>
                                        <div class="product__block--cost"><?= $row['cost']; ?></div>
                                        </div>
                                        <button class="product__block--info--like">
                                            <img src="images/Header/heart .svg" width="30" height="30" alt="">
                                        </button>
                                    </div>
    
                                    <div class="product__block--buy">
                                        <button class="product__buy--button">Купить</button>
                                        
                                    </div>
                                
                            </div>
                        
                        
                        
                        
                        
                        
                        <?php }
                        
                        ?>

Вот код connect.php

<?php
$connect = mysqli_connect('localhost','root','root','registerAndLogin');
if (!$connect){
    echo "Ошибка - ".mysqli_connect_error();
}

?>

Answer 1

[Spartak (Web-StyleStudio)]

while($row = mysqli_fetch_array($result))

while($row = mysqli_fetch_array($results))
Одну букву пропустили в $results

Comments

[Stoble]

Под вечер уже совсем невнимательным становлюсь, спасибо

[Spartak (Web-StyleStudio)]

Stoble,
Бывает))