Ошибка mysqli_fetch_array() expects parameter 1 to be mysqli_result

Question

Сам запрос рабочий. Выводит напрямую из БД то что нужно, но почему-то на странице ошибку выдает. Кто-нибудь знает,как это можно исправить? Весь код страницы:

    <?php
        $page = isset($_GET["page"]) ? (int) $_GET["page"] : 1;
    $notesOnPage = 3;

    $from = ($page - 1) * $notesOnPage;

    $today = date(&quot;F j, Y, g:i a&quot;);  
    $query = &quot;SELECT `discription`, `author`, `data`, `tags`.`Name` FROM `quote`, `tags` ORDER BY id DESC LIMIT $from, $notesOnPage&quot;;
    var_dump($query);
    $sql = mysqli_query($connect, $query);
    $number = 0;

    while ($news = mysqli_fetch_array($sql))
    {
?&gt;

&lt;div class=&quot;card news border-dark mb-3&quot; style=&quot;min-width: 100%;&quot;&gt;
    &lt;div class=&quot;card-header&quot;&gt;Quotes&lt;/div&gt;
        &lt;div class=&quot;card-body text-dark&quot;&gt;
            &lt;input type=&quot;checkbox&quot; class=&quot;read-more-checker&quot; id=&quot;read-more-checker-&lt;?php echo $number; ?&gt;&quot; /&gt;
            &lt;div class=&quot;limiter&quot;&gt;                   
                &lt;p class=&quot;card-text&quot;&gt;Автор: &lt;?php echo $news['author']; ?&gt;&lt;/p&gt;
                &lt;p class=&quot;card-text&quot;&gt;&lt;?php echo $news['discription']; ?&gt;&lt;/p&gt;
                &lt;p class=&quot;card-text&quot;&gt;&lt;?php echo $news['data']; ?&gt;&lt;/p&gt;
                &lt;div class=&quot;bottom&quot;&gt;&lt;/div&gt;
            &lt;/div&gt;
            &lt;label for=&quot;read-more-checker-&lt;?php echo $number; ?&gt;&quot; class=&quot;read-more-button&quot;&gt;&lt;/label&gt;
        &lt;/div&gt;  
&lt;/div&gt;


<?php
        $number = $number + 1;
    }
$select = mysqli_query($connect, &quot;SELECT COUNT(*) FROM `quote`&quot;);
if (!$select) die ('Error'. mysqli_error($connect));
$row = mysqli_fetch_array($select);
$count_post = $row [0];

$pages_count = ceil ($count_post / $notesOnPage);
//var_dump($pages_count);

?>
    <div class="pagination">
<?php
for($i = 1; $i &lt; $pages_count; $i++){
    echo &quot;&lt;a href =\&quot;?page=$i\&quot;&gt;$i &lt;/a&gt;&quot;;
}

?>
</div>
<?php
    include_once("footer.php");
?>

Ошибка: Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, bool given in C:\xampp\htdocs\quotes\index.php on line 23

Линия 23 - while ($news = mysqli_fetch_array($sql))