我正在尝试在此代码中使用 echo

<?php
                            
    require 'connect.php';
    $query = mysqli_query($conn, "SELECT * FROM `videos`") or die(mysqli_error());
    while($fetch = mysqli_fetch_array($query)){
           $id= $fetch['id'];
?>
    <?php echo "
        <li class='col-lg-4 col-md-6 col-dm-12'>
        <div class='da-card box-shadow'>
        <div class='da-card-photo'>
            <img src=' echo $fetch['video_name']' alt=''>
            <div class='da-overlay'>
                <div class='da-social'> 
                    <h4 class='mb-10 color-white pd-20'></h4>
                    <ul class='clearfix'>
                        <li>
                            <a href='' data-fancybox='images'><i class='fa fa-picture-o'></i></a>
                        </li>
                        <li>
                            <a href='#'><i class='fa fa-link'></i></a>
                        </li>
                    </ul>
                </div>
            </div>
        </div>
    </div>
        </li>
    "; ?>
    <?php
    }
?>

我收到此错误 Parse 错误：语法错误，意外的字符串内容“”，期望“-”或标识符或变量或数字位于 C:\xampp\htdocs\desk\gallery.php 行 572 为什么会出现这个错误？ ...................................................... …………