How can I create a link to download a telegram file?

Question

Python

Here is all the data I have:

[{'_': 'Message', 'id': 799, 'peer_id': {'_': 'PeerChannel', 'channel_id': 1195851415}, 'date': datetime.datetime(2021, 5, 25, 18, 16, 29, tzinfo=datetime.timezone.utc), 'message': '', 'out': True, 'mentioned': False, 'media_unread': False, 'silent': True, 'post': True, 'from_scheduled': False, 'legacy': False, 'edit_hide': False, 'pinned': False, 'from_id': None, 'fwd_from': None, 'via_bot_id': None, 'reply_to': None, 'media': {'_': 'MessageMediaDocument', 'document': {'_': 'Document', 'id': 5289805549955714500, 'access_hash': 6708206575358420852, 'file_reference': b'\x04GG>\x97\x00\x00\x03\x1f`\xad?\x1a\xc1\xbc\x8c\xba%\x82J\xa9\xa8\xd4\xa7\xd2\xa1\xd29\xde', 'date': datetime.datetime(2021, 5, 25, 18, 16, 29, tzinfo=datetime.timezone.utc), 'mime_type': 'video/mp4', 'size': 4058427, 'dc_id': 2, 'attributes': [{'_': 'DocumentAttributeVideo', 'duration': 10, 'w': 576, 'h': 1024, 'round_message': False, 'supports_streaming': True}], 'thumbs': [{'_': 'PhotoStrippedSize', 'type': 'i', 'bytes': b'\x01(\x16\xc9g\xdc1\xb5G\xd0S)\xeb\x19\xcf\xcc\x08\xfaR\xf9c=[\x1f\xee\xff\x00\xf5\xe9\xd9\xf6\x0b\xa2:)\xc5\x0ex\x07\x14Q\xca\xc5tXQV\x1fo\x97\x8d\xc4\xfau\xac\xfd\xed\xebG\x98\xff\x00\xde5\xa4\xa6\x9bD\xa8\xb4Nh\xa8\x0c\x8c\x7f\x8a\x8a\xafj\x83\x94e\x14QX\x16\x14QE\x00\x7f'}, {'_': 'PhotoSize', 'type': 'm', 'location': {'_': 'FileLocationToBeDeprecated', 'volume_id': 200281400252, 'local_id': 16815}, 'w': 180, 'h': 320, 'size': 7162}], 'video_thumbs': []}, 'ttl_seconds': None}, 'reply_markup': None, 'entities': [], 'views': 1, 'forwards': 0, 'replies': None, 'edit_date': None, 'post_author': 'Adrian', 'grouped_id': None, 'restriction_reason': [], 'ttl_period': None}]

How to collect a link to download the video from this?

Answer 1

The link will not work, but you can take the file through the client API. Request from telethon, so we use it.

location = types.InputDocumentFileLocation(id=5289805549955714500, access_hash=6708206575358420852, file_reference=b'\x04GG>\x97\x00\x00\x03\x1f`\xad?\x1a\xc1\xbc\x8c\xba%\x82J\xa9\xa8\xd4\xa7\xd2\xa1\xd29\xde', thumb_size='100')

with open('result.mp4', 'wb') as f:
        client.download_file(input_location=location, file=f)

Something like this. But this is in case the data is really presented in such a dictionary, and this is not a class printout from the code. If there is an object with a message in the code, then there is no need to fence anything, and then

with open('result.mp4', 'wb') as f:
        msg.download_media(file=f)