Question

<?php

while($row=mysql_fetch_array($result))
{ 
   if(in_array($ext,$audioArr))
   { 
    $aud=$row['news_media'];
?>
    <div id="player2" class="aplayer"></div> 

    <script>
    var ap2 = new APlayer({
        element: document.getElementById('player2'),
        narrow: true,
        autoplay: false,
        showlrc: false,
        music: {
            title: 'Preparation',
            author: 'Hans Zimmer/Richard Harvey',
            url: 'media/<?php echo $row['news_media']; ?> ',
            pic: 'icons/2.jpg'
        }
    });
    ap2.init();  
    </script>
    <?php
    }
}?>

在我的代码中，我有一个重复打印音频的循环。但我不知道如何在js中打印php变量echo $row['news_media']。我想在url打印这个：'media /？＆gt; ”。所以请帮助我。我从这个网站尝试了很多技巧，但我找不到合适的技巧。

Answer 1

我想你错过了这个：

    <?php

while($row=mysql_fetch_array($result))
{ 
   if(in_array($ext,$audioArr))
   { 
    $aud=$row['news_media'];
    ?> // This close
    <div id="player2" class="aplayer"></div> 

    <script>
    var ap2 = new APlayer({
        element: document.getElementById('player2'),
        narrow: true,
        autoplay: false,
        showlrc: false,
        music: {
            title: 'Preparation',
            author: 'Hans Zimmer/Richard Harvey',
            url: 'media/<?php echo $row['news_media']; ?> ',
            pic: 'icons/2.jpg'
        }
    });
    ap2.init();  
    </script>
    <?php
    }
}?>

Answer 2

替换

url: 'media/<?php echo $row['news_media']; ?> ',

带

url: 'media/'+<?php echo $row['news_media']; ?>,

希望这会有所帮助。

享受！

你在php close标签中有问题。以下是解决方案。试试吧

<?php

while($row=mysql_fetch_array($result))
{ 
   if(in_array($ext,$audioArr))
   { 
    $aud=$row['news_media'];
?>
    <div id="player2" class="aplayer"></div> 

    <script>
    var ap2 = new APlayer({
        element: document.getElementById('player2'),
        narrow: true,
        autoplay: false,
        showlrc: false,
        music: {
            title: 'Preparation',
            author: 'Hans Zimmer/Richard Harvey',
            url: 'media/'+<?php echo $row['news_media']; ?>,
            pic: 'icons/2.jpg'
        }
    });
    ap2.init();  
    </script>
    <?php
    }
}?>

在js

2 个答案: