Based on your search, it seems you are looking for either that already does this, or how to build your own .
# Run download in a thread to avoid blocking def job(): msg = bot.send_message(message.chat.id, "Starting download...") try: with tempfile.TemporaryDirectory() as td: if choice.lower() == 'all': # Download whole playlist info = download_video(url, td, audio_only=(fmt=='mp3')) # Find downloaded files and send as individual messages or as zip files = [] for root, _, filenames in os.walk(td): for fn in filenames: files.append(os.path.join(root, fn)) # If many files, consider zipping if len(files) > 10 or sum(os.path.getsize(f) for f in files) > 50*1024*1024: import zipfile zip_path = os.path.join(td, 'playlist.zip') with zipfile.ZipFile(zip_path, 'w') as z: for f in files: z.write(f, os.path.basename(f)) bot.send_document(message.chat.id, open(zip_path, 'rb')) else: for f in files: bot.send_document(message.chat.id, open(f, 'rb')) else: # Single index try: idx = int(choice) except: bot.send_message(message.chat.id, "Index must be a number or 'all'.") return # Use yt-dlp playlist item syntax: url&index=X or "https://www.youtube.com/watch?v=id" # Simplest: download playlist but instruct to download single entry via playlist indices ytdl_url = url + f"&index=idx" info = download_video(ytdl_url, td, audio_only=(fmt=='mp3')) # Send files in td for root, _, filenames in os.walk(td): for fn in filenames: path = os.path.join(root, fn) bot.send_document(message.chat.id, open(path, 'rb')) bot.edit_message_text("Download completed.", chat_id=message.chat.id, message_id=msg.message_id) except Exception as e: bot.send_message(message.chat.id, f"Error: e") telegram bot to download youtube playlist hot
When using a Telegram bot to download YouTube playlists, there are a few safety and security considerations to keep in mind: how to find a popular ("hot") bot Based