Хотел сделать поиск по сайту, mysqli_fetch_assoc () ожидает, что параметр 1 будет mysqli_result, логическое значение дано в

Question

<div class="nav__common d-flex justify-content-between m-5">
        <nav class="nav"><img class="size" src="img/logo1.png" alt=""/></nav>
        <form method="POST">
          <input type="text" name="search" class="search"><input type="submit" name="submit" value="поиск"> 
        </form>
        <nav class="nav"><a class="nav-link text-body" href="index.html">Главный</a><a class="nav-link text-body" href="work.html">Каталог</a><a class="nav-link text-body" href="blog.html">Отзывы</a><a class="nav-link border btn btn-outline-primary rounded-pill border-primary text-bodyp-5" href="log.php">Регистрация</a></nav>
      </div>
      <hr>
      <?php
      if(isset($_POST['submit'])){
        $search = $_POST['search'];
        $sql = "SELECT * FROM products WHERE `title` LIKE `%$search%` OR `opisanie` LIKE `%$search%` ";
        $query =mysqli_query($link, $sql);
        while($row = mysqli_fetch_assoc($query)){ echo '<h1>'.$row ['title'].'</h1><p>' .$row['opisanie']. '</p><br>';}
      } 
      ?>