如何使用Python子进程添加到VLC播放列表队列

Question

我尝试使用python2.7subprocess库以编程方式将歌曲添加到VLC播放器队列中。在

从here和{a2}，我可以启动VLC播放器并播放一首歌（或从一开始就对歌曲进行排队）

from subprocess import Popen
vlcpath = r'C:\Program Files (x86)\VideoLAN\VLC\vlc.exe'
musicpath1 = r'path\to\song1.mp3'
musicpath2 = r'path\to\song2.mp3'
p = Popen([vlcpath,musicpath1]) # launch VLC and play song
p = Popen([vlcpath,musicpath1,musicpath2]) # launch VLC and play/queue songs

问题是我不知道启动时的整个队列播放列表。我希望能够将歌曲添加到已经运行的VLC进程的队列中。请问我该怎么做？在

从here开始，我认为合适的命令行条目是：

^{pr2}$

但我不知道在subprocess中执行此操作的语法。我试了几件事，但都没用：

• 再次调用Popen（打开一个新进程）
• 调用p.communicate（我以为这是如何输入stdin命令的）

Answer 1

要在Windows上运行命令：vlc.exe started-from-file playlist-enqueue "2.wmv" 使用subprocess模块：

from subprocess import Popen

cmd = 'vlc.exe  started-from-file  playlist-enqueue "2.wmv"'
p = Popen(cmd) # start and forget
assert not p.poll() # assert that it is started successfully

要等待命令完成：

^{pr2}$

---

But how do I run that command a second time on the same p process? Your code starts a new instance of VLC, rather than running that on top of the p that was already open. I found that if I run the vlc.exe started-from-file playlist-enqueue "2.wmv" command multiple times manually (in a command prompt window), it correctly launches vlc (the first time) and then adds to queue (on subsequent calls). So I think I just need to be able to run the code you suggested multiple times "on top of itself"

每个Popen()都会启动一个新的进程。每次在命令行中手动运行该命令时，它都会启动一个新进程。它可能取决于系统上当前的vlc配置，无论它保留多个vlc实例，还是运行不同的命令（不同的命令行参数）。在