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fix: handle /help command directly in Telegram, bypassing ACL check

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The /help command was routed through _forward_command → _handle_message
→ is_allowed(), which denied access to users not in the allowFrom list.
Since /help is purely informational, it should be accessible to all
users — similar to how /start already works with its own handler.

Add a dedicated _on_help handler that replies directly without going
through the message bus access control.

Closes #687
This commit is contained in:
Nikolas de Hor
2026-02-18 21:27:13 -03:00
parent 8de36d398f
commit 33d760d312
1 changed files with 17 additions and 3 deletions
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20
nanobot/channels/telegram.py
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@@ -146,7 +146,7 @@ class TelegramChannel(BaseChannel):
# Add command handlers
self._app.add_handler(CommandHandler("start", self._on_start))
self._app.add_handler(CommandHandler("new", self._forward_command))
self._app.add_handler(CommandHandler("help", self._forward_command))
self._app.add_handler(CommandHandler("help", self._on_help))
# Add message handler for text, photos, voice, documents
self._app.add_handler(
@@ -258,14 +258,28 @@ class TelegramChannel(BaseChannel):
"""Handle /start command."""
if not update.message or not update.effective_user:
return
user = update.effective_user
await update.message.reply_text(
f"👋 Hi {user.first_name}! I'm nanobot.\n\n"
"Send me a message and I'll respond!\n"
"Type /help to see available commands."
)
async def _on_help(self, update: Update, context: ContextTypes.DEFAULT_TYPE) -> None:
"""Handle /help command directly, bypassing access control.
/help is informational and should be accessible to all users,
even those not in the allowFrom list.
"""
if not update.message:
return
await update.message.reply_text(
"🐈 nanobot commands:\n"
"/new — Start a new conversation\n"
"/help — Show available commands"
)
@staticmethod
def _sender_id(user) -> str:
"""Build sender_id with username for allowlist matching."""