I am trying to make a program in python to open a VLC "Browsing File window" on Ubuntu
global process
import io,sys,os
import subprocess
myprocess = subprocess.call(['vlc','/home/tansen'])
but the code above just opens the 'VLC Player' not the file opening window
Can you please guide me on how to get the required result
I am adding the vlc filing opening image as well
Thank you
according to the documentation, the correct syntax should be
vlc -vvv video.mp4
and the python code that you can use is
subprocess.POPEN(['vlc', '-vvv', '/path/to/video.mp4'])
You can also add the PIPE to dump the output from vlc as well.
What OS platform?
On Linux: If you run qdbusviewer you'll see the dbus methods available. I don't see one for showing the file dialog, but there is one for opening a URL:
qdbus org.mpris.MediaPlayer2.vlc /org/mpris/MediaPlayer2 org.mpris.MediaPlayer2.Player.OpenUri test.mp3
So from Python:
import subprocess
subprocess.call(["qdbus", "org.mpris.MediaPlayer2.vlc", "/org/mpris/MediaPlayer2", "org.mpris.MediaPlayer2.Player.OpenUri", "file:///home/john/test.mp3"])'
Or:
import gobject
gobject.threads_init()
from dbus import glib
glib.init_threads()
import dbus
bus = dbus.SessionBus()
obj = bus.get_object("org.mpris.MediaPlayer2.vlc", "/org/mpris/MediaPlayer2")
iface = dbus.Interface(obj, "org.mpris.MediaPlayer2.Player")
iface.OpenUri("file:///home/john/test.mp3")
On Windows: Try COM?
