我的网站有问题,我使用JavaScript来显示自定义的视频播放器。我从数据库中加载了几个视频,并连续显示它们,问题是JavaScript仅适用于第一个视频。我知道我将不得不更改例如变量名以使脚本多次运行,但是是否有另一种方式可以为每个videoelement加载相同的脚本?
这是我的代码,可以更好地理解:
while ($row = mysqli_fetch_assoc($result))
{
$location = "../" . $row['pfad'] . $row['video'];
$id = $row['id'];
?>
<br>
<p><?php echo "$row[titel]"; ?></p>
<div class="video-container">
<div class="c-video">
<video id="video" class="video" src="<?php echo "$location"; ?>"></video>
<div class="controls">
<div id="process-bar" class="process-bar">
<div id="currentBuffer" class="currentBuffer"></div>
<div id="currentProcess" class="currentProcess"></div>
</div>
<div class="buttons">
<button id="skip-back" class="skip-back" data-skip="-10"></button>
<button id="play-pause"></button>
<button id="skip-front" class="skip-front" data-skip="10"></button>
<span id="videoCurrentTime" class="videoTimer"></span> <span class="videoTimer">/</span> <span id="videoTime" class="videoTimer"></span>
<button id="sound" class="highSound"></button>
<input type="range" name="volume" class="volume-slider" id="volume-slider" min="0" max="1" value="1" step="0.05">
<button id="fullscreen" class="fullscreen"></button>
</div>
</div>
</div>
<script type="text/javascript" src="../javascript/videoplayer.js"></script>
</div>
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