Question

我将歌曲名称传递给某个功能，并且无法为omxplayer构建路径

def PlayMusic(song):
  #check if the process exists
  is_pid = subprocess.call("pidof omxplayer.bin > /dev/null", shell=True)
  if is_pid == 0:
    return musicplaying("Song already Playing")
  else:
    pathsong ='/home/pi/'+song
    os.system('omxplayer --no-keys -o local pathsong &')
    return musicplaying (song +" playing")

我如何把＆＃39;途径＆＃39;进入omxplayer命令行？

Answer 1

这就是我想要的

<!DOCTYPE html>
<html>
<head>
    <title><?php echo get_bloginfo( 'name' ); ?> - <?php bloginfo('description'); ?></title>

    <link rel="stylesheet" type="text/css" href="<?= get_template_directory_uri(); ?>/css/bootstrap.css">
    <link rel="stylesheet" type="text/css" href="<?= get_template_directory_uri(); ?>/css/style.css">



</head>
<body>
<nav class="navbar navbar-default navbar-inverse navbar-fixed-top" role="navigation">
   <div class="container-fluid" id="navfluid">
       <div class="navbar-header">
           <button type="button" class="navbar-toggle" data-toggle="collapse" data-target="#navigationbar">
               <span class="sr-only">Toggle navigation</span>
               <span class="icon-bar"></span>
               <span class="icon-bar"></span>
               <span class="icon-bar"></span>
            </button>
            <a class="navbar-brand" href="<?php echo get_home_url() ?>"><?php echo get_bloginfo( 'name' ); ?></a>
       </div>
       <div class="collapse navbar-collapse" id="navigationbar">
           <ul class="nav navbar-nav">
               <li><a href="<?php echo get_home_url() ?>"><span class="glyphicon glyphicon-home" aria-hidden="true"></span> Home</a></li>
        <li><a href="news"><span class="glyphicon glyphicon-pencil" aria-hidden="true"></span> Blog </a></li>

           </ul>
      </div>
   </div>
</nav>

在python中将歌曲名称传递给omxplayer

1 个答案: