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• Elegant early exit from a spawned ffmpeg process in C

  21 octobre 2016, par jackson80

  I have a program where I build an ffmpeg command string to capture videos with options input through a gtk3 gui. Once I have all my options selected, I spawn a process with the ffmpeg command string. And I add a child watch to tell me when the process has completed.

    // Spawn child process 
  
    ret = g_spawn_async (NULL, argin, NULL, G_SPAWN_DO_NOT_REAP_CHILD, NULL, NULL, &pid1, NULL);
  
    if ( !ret )
  
    {
  
      g_error ("SPAWN FAILED");
  
      return;
  
    }
  
  
  
  /* Add watch function to catch termination of the process.  This function
  
     * will clean any remnants of process */
  
    g_child_watch_add (pid1, (GChildWatchFunc)cb_child_watch, widget );

  Executing ffmpeg from a terminal using a command line, the program will give an option to input a "q" at the terminal to end the ffmpeg process early.
  Is there any way to send a "q" to that spawned process to elegantly end the ffmpeg ? I’m fairly sure I could kill the process using the process id, but I would rather stop it using a mechanism that allows ffmpeg to gracefully exit..
  This is running Centos 7, kernel 4.7.5, ffmpeg version 3.0.2.
  Since I can still access the terminal where the ffmpeg output is displayed, I’ve tried typing a "q", but it has no effect on the process.

• AttributeError : 'FFmpegAudio' object has no attribute '_process' while trying to play audio from URL

  13 août 2022, par jmcamacho7

  I can't find any solution online and I don't know what's wrong.

  


  My code is : (Not pasting the URL getting since that works fine)

  


  from urllib import parse, request
import re
import pafy
from discord import FFmpegPCMAudio, PCMVolumeTransformer

FFMPEG_OPTIONS = {
    'before_options': '-reconnect 1 -reconnect_streamed 1 -reconnect_delay_max 5', 'options': '-vn'
}

@bot.command(pass_context=True)
async def play(ctx, * , search):
  query_string = parse.urlencode({'search_query': search})
  html_content = request.urlopen('http://www.youtube.com/results?' + query_string)
  search_results=re.findall('watch\?v=(.{11})',html_content.read().decode('utf-8'))
  print(search_results[0])
  
  if(ctx.author.voice):
    channel = ctx.message.author.voice.channel
    await ctx.send("https://www.youtube.com/watch?v="+search_results[0])  
    url = "https://www.youtube.com/watch?v="+search_results[0]
    conn = await channel.connect()
    conn.play(discord.FFmpegAudio(url, **FFMPEG_OPTIONS))
  else:
    await ctx.send("Necesitas estar en un canal de audio para usar este comando")


  


  It just gives me this error everytime I try it :

  


  Traceback (most recent call last):
  File "/home/runner/HakuBot/venv/lib/python3.8/site-packages/discord/player.py", line 103, in __del__
    self.cleanup()
  File "/home/runner/HakuBot/venv/lib/python3.8/site-packages/discord/player.py", line 154, in cleanup
    proc = self._process
AttributeError: 'FFmpegAudio' object has no attribute '_process'


  


  Anyway to solve this ?

  


• ffmpeg : Get last 10 seconds of live stream [duplicate]

  7 janvier 2020, par Manuel

  This question already has an answer here :

  • FFMpeg Copy Live Stream (Limit to 60s file)
    
    2 answers
    

  I am having a live stream from an ip camera, where I would like to get only the last 10 seconds of that stream.
  I’ve tried parse the steam continuously into an .mp4, and use ffmpeg -sseof afterwards, but this fails, as the input file is still writing.

  Has anyone an idea, if this is homehow possible ?

  Thanks !