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Hi all:

Apologize for basic question, I am new at this kind of development. I am trying to get VLc To work using python scripting.
I downloaded python 3.5.1, and the vlc.py file
how do I get started to launch vlc from python?

>>> import vlc
>>> screen1 = vlc.Instance(st)
Traceback (most recent call last):
File "<pyshell#7>", line 1, in <module>
screen1 = vlc.Instance(st)
NameError: name 'st' is not defined
>>> screen1 = vlc.Instance()
>>> player1=screen1.media_player_new()
>>> player1.set_media()
Traceback (most recent call last):
File "<pyshell#10>", line 1, in <module>
player1.set_media()
TypeError: set_media() missing 1 required positional argument: 'p_md'
>>> media1=screen1.media_new('c:\\VLCTest\Video-1028.avi')
>>> media1
<vlc.Media object at 0x0333C8B0>
>>> print media1
SyntaxError: Missing parentheses in call to 'print'
>>> print (media1)
<vlc.Media object at 0x0333C8B0>
>>> player1.set_media(media1)
>>> player1.play()
0
>>>


It does not launch a new instance of Media player, how can I force it to bring up the Media player from the python script?
ANy thing in setup I am missing?
Can anyone help? thanks

Thanks

Hi,

I am also new to this stuff, but I think you misunderstand this module. It does not start the gui executable. You have to program the gui and connect a window handle of it with your player object. Look into example tkvlc.py class Player methods __init__ and OnOpen.

In fact (on windows and linux at least, macosx is different in this respect), if no video widget is specified, a new one will be created by libvlc. So the above code should work. Note that it could be shortened to If you do not see any output, you can enable more verbose messages to have an idea of the issue:

oh, great, then, I guess, I can do without a working log_set()