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Javascript 消失或隐藏后停止歌唱?_Javascript_Jquery_Playback - Fatal编程技术网

Javascript 消失或隐藏后停止歌唱?

Javascript 消失或隐藏后停止歌唱?,javascript,jquery,playback,Javascript,Jquery,Playback,我正在开发一个由HTML扩展支持的音乐 一旦隐藏,我如何阻止歌曲播放();或衰减();函数被调用了吗 这里是HTML和PHP(是的,安全问题,仍处于开发阶段,所以我将修复所有安全性和bug,一旦我 <?php require "core/database.php"; $id = $_GET['id']; $select_song = mysql_query("SELECT * FROM music_player WHERE song_id='$id'") or

我正在开发一个由HTML扩展支持的音乐

一旦隐藏,我如何阻止歌曲播放();或衰减();函数被调用了吗

这里是HTML和PHP(是的,安全问题,仍处于开发阶段,所以我将修复所有安全性和bug,一旦我

<?php
    require "core/database.php";

    $id = $_GET['id'];

    $select_song = mysql_query("SELECT * FROM music_player WHERE song_id='$id'") or die("we got an error");
    $retrive_song = mysql_fetch_array($select_song);
    $song_id = $retrive_song['song_id'];
    $song_path = $retrive_song['song_path'];
    $song_name = $retrive_song['song_name'];
    $band = $retrive_song['band'];
    echo($song_name);


    echo "
    <html>
    <head>
    </head>
    <script src='js/jquery.js'></script>
    <script src='js/custom.js'></script>

        <div id='content' style='display:none;'>
    <audio style=' display:block;' controls>

      <source src='music/$song_path'>
    Your browser does not support the audio element.
    </div>
    </audio>
        <b>Other Songs From $band</b>

    ";
        $select_band = mysql_query("SELECT * FROM music_player WHERE band='$band'");

        while($get_band = mysql_fetch_array($select_band)) {

        $band_song = $get_band['song_name'];
        if($song_name!=$band_song) {
    echo "
    <p> 

    <a href='#' class='songs_link'> $band_song </a> <br>

    </p>
    </html>


    ";
    }
    }

?>

我正在尝试淡出当前歌曲,然后在单击链接后转到用户选择的新歌。

向音频元素添加id,并:

(...).fadeOut(function(){
   $('#audio_element').get(0).pause(); // or .stop()
});
暂停有效,但将其从暂停更改为停止时,停止()功能仍不起作用
(...).fadeOut(function(){
   $('#audio_element').get(0).pause(); // or .stop()
});