No longer run send_code_request from sign_in

readthedocs/misc/v2-migration-guide.rst

@@ -428,6 +428,13 @@ However, most likely, you were already doing the right thing (or else you would'
 "why is this not being edited", which you would most likely consider a bug rather than a feature).
 
 
+Signing in no longer sends the code
+-----------------------------------
+
+``client.sign_in()`` used to run ``client.send_code_request()`` if you only provided the phone and
+not the code. It no longer does this. If you need that convenience, use ``client.start()`` instead.
+
+
 The client.disconnected property has been removed
 -------------------------------------------------
 

telethon/_client/auth.py

@@ -225,9 +225,7 @@ async def sign_in(
     if me:
         return me
 
-    if phone and not code and not password:
-        return await self.send_code_request(phone)
-    elif code:
+    if phone and code:
         phone, phone_code_hash = \
             _parse_phone_and_hash(self, phone, phone_code_hash)
 
@@ -247,10 +245,7 @@ async def sign_in(
             api_id=self.api_id, api_hash=self.api_hash
         )
     else:
-        raise ValueError(
-            'You must provide a phone and a code the first time, '
-            'and a password only if an RPCError was raised before.'
-        )
+        raise ValueError('You must provide either phone and code, password, or bot_token.')
 
     result = await self(request)
     if isinstance(result, _tl.auth.AuthorizationSignUpRequired):

telethon/_client/telegramclient.py

@@ -381,8 +381,6 @@ class TelegramClient:
 
         You should only use this if you are not authorized yet.
 
-        This method will send the code if it's not provided.
-
         .. note::
 
             In most cases, you should simply use `start()` and not this method.