MySQLifetch_all的奇怪错误-WeirderrorwithMySQLifetch_all

This is probably going to be marked as a duplicate but my error seems to be different from the others that I've seen here on stackoverflow.

这可能会被标记为重复，但我的错误似乎与我在stackoverflow上看到的其他错误不同。

the following code is supposed to display the content from the image: code:

以下代码应该显示图像中的内容：代码：

 if($news_result = $db->query("SELECT * FROM `news` ORDER BY `likes` DESC LIMIT 4")) {                      
    if ($news_result->num_rows) {
        $rows = $news_result->fetch_all(MYSQL_ASSOC);
            $row_count = 0;
            foreach($rows as $row) {
                $row_count++;

                $content_echo = '
                    

                    
                    
    
                            
'.$row['title'].'


                            

                                

                                    

                                        
'.$row['description'].'
                                    
                                
                                

                                    

                                        
BY: '.$row['creator'].'
                                    
                                
                            
                            
'.$row['likes'].'
                            

                                
READ MORE
  
                            
                        
                    
                
';

                $separator = '  

                

                
';

                if ($news_result->num_rows > 1 && $row_count > 0 && $row_count <$news_result->num_rows) {
                    $content_echo .= $separator;
                }
                echo $content_echo;
            }
        }
    }

Image:

'.$row['title'].'