在页面上运行该页面时,页面加载和暂停需要很长时间,然后一次显示所有内容,我有什么方法可以斋戒?
我尝试过的操作是将该部分保存为.html并从那里显示仍然没有什么不同,还尝试使用lozad.js加载图像仍然可以做到这一点
<div class='person-videos-text'>
<div class="row">
<?php
while($row = mysqli_fetch_assoc($recentTreningVideos))
{
$TrendImageURL = $row['youtubethumbnail'];
$TrendSongName = $row['musicvideotitle'];
$TrendCleanSongname = $row['songname'];
$name = $row['youtubethumbnail'];
$rippedUrl = explode("/",$name);
//print_r($rippedUrl);
$hostt = $rippedUrl[0];
$hostUrl = $rippedUrl[2];
$imageNameVi = $rippedUrl[3];
$imageNameUrl = $rippedUrl[4];
$imageType = $rippedUrl[5];
?>
<div class="column trending">
<div class="card">
<a href="viewactors.php?songname=<?php echo $TrendCleanSongname; ?>">
<img class="trending-image-size" src="cached/<?php echo $imageNameUrl;?>/<?php echo $imageType; ?>" alt="John">
<div class="person-container">
<!--<h2>John Doe</h2>-->
<p class="title trending-title"><?php echo $TrendSongName; ?></p>
<!--<p>Some text that describes me lorem ipsum ipsum lorem.</p>
<p>example@example.com</p>
<p><button class="button">Contact</button></p>-->
</div>
</a>
</div>
</div>
<?php } ?>
</div>
</div>