Closed undaunt closed 3 years ago
Make a series folder with playlist Id per playlist
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I replied to https://github.com/ZeroQI/YouTube-Agent.bundle/issues/59 regarding this line of thinking.
2021-01-26 16:55:30,615 (7faa757fa700) : INFO (init:489) - Grouping folder not found or single folder, root: /data/youtube/Dead Suns Character Intros - [PLz3Be--ot61Mvmms68yBEhX6KrzPrqxsW], path: , Grouping folder: , subdirs: 0, reverse_path: [''] 2021-01-26 16:55:30,616 (7faa757fa700) : INFO (init:575) - No GUID so random folder
It seem like the Playlist ID is not recognized despite being correct (tested it)
https://github.com/ZeroQI/YouTube-Agent.bundle/blob/master/Contents/Code/__init__.py line 697
YOUTUBE_REGEX_PLAYLIST = Regex('[(?:youtube-)?(?P
Please replace it with the below, update metadata for this series, and report, and will add i to the master
YOUTUBE_REGEXPLAYLIST = Regex('[(?:youtube-)?(?P
Is your feature request related to a problem? Please describe. I'm not currently able to figure out how to do this, so I apologize if it is already possible. My goal is to download a given YouTube channel's playlists into individual season folders. I'm currently causing all my files to either show up as one large season, or by year seasons inadvertently.
Describe the solution you'd like Here is how I'm currently trying to organize the data.
Playlists %(uploader)s - [%(uploader_id)s]/Playlist - [%(playlist_id)s]/%(playlist_index)s - %(title)s [%(id)s].%(ext)s
Non-Playlist Videos (to be downloaded after all playlist videos, skipping them by using youtube-dl's archive file) %(uploader)s - [%(uploader_id)s]/No Playlist/%(upload_date)s - %(title)s [%(id)s].%(ext)s
Please let me know if the prior could be improved upon. I'm currently using the Absolute Series Scanner and the Youtube bundle, both latest versions as of today.
Describe alternatives you've considered An alternative could be each playlist being a series instead of a season, and using a collection to view all playlists and videos belonging to a channel. I'm not sure how to accomplish that either.