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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