Get first association from a has many through association - ruby-on-rails

Question

I'm trying to join the first song of each playlist to an array of playlists and am having a pretty tough time finding an efficient solution.
I have the following models:
class Playlist < ActiveRecord::Base
belongs_to :user
has_many :playlist_songs
has_many :songs, :through => :playlist_songs
end
class PlaylistSong < ActiveRecord::Base
belongs_to :playlist
belongs_to :song
end
class Song < ActiveRecord::Base
has_many :playlist_songs
has_many :playlists, :through => :playlist_songs
end
I would like to get this:
playlist_name | song_name
----------------------------
chill | baby
fun | bffs
I'm having a pretty tough time finding an efficient way to do this through a join.
UPDATE ****
Shane Andrade has lead me in the right direction, but I still can't get exactly what I want.
This is as far as I've been able to get:
playlists = Playlist.where('id in (1,2,3)')
playlists.joins(:playlist_songs)
.group('playlists.id')
.select('MIN(songs.id) as song_id, playlists.name as playlist_name')
This gives me:
playlist_name | song_id
---------------------------
chill | 1
This is close, but I need the first song(according to id)'s name.

Answer 1

Assuming you are on Postgresql
Playlist.
select("DISTINCT ON(playlists.id) playlists.id,
songs.id song_id,
playlists.name,
songs.name first_song_name").
joins(:songs).
order("id, song_id").
map do |pl|
[pl.id, pl.name, pl.first_song_name]
end

Answer 2

I think this problem would be improved by having a a stricter definition of "first". I'd suggest adding a position field on the PlaylistSong model. At which point you can then simply do:
Playlist.joins(:playlist_song).joins(:song).where(:position => 1)

Answer 3

What you are doing above with joins is what you would do if you wanted to find every playlist with a given name and a given song. In order to collect the playlist_name and first song_name from each playlist you can do this:
Playlist.includes(:songs).all.collect{|play_list| [playlist.name, playlist.songs.first.name]}
This will return an array in this form [[playlist_name, first song_name],[another_playlist_name, first_song_name]]

Answer 4

I think the best way to do this is to use an inner query to get the first item and then join on it.
Untested but this is the basic idea:
# gnerate the sql query that selects the first item per playlist
inner_query = Song.group('playlist_id').select('MIN(id) as id, playlist_id').to_sql
#playlists = Playlist
.joins("INNER JOIN (#{inner_query}) as first_songs ON first_songs.playlist_id = playlist.id")
.joins("INNER JOIN songs on songs.id = first_songs.id")
Then rejoin back to the songs table since we need the song name. I'm not sure if rails is smart enough to select the song fields on the last join. If not you might need to include a select at the end that selects playlists.*, songs.* or something.

Answer 5

Try:
PlaylistSong.includes(:song, :playlist).
find(PlaylistSong.group("playlist_id").pluck(:id)).each do |ps|
puts "Playlist: #{ps.playlist.name}, Song: #{ps.song.name}"
end
(0.3ms) SELECT id FROM `playlist_songs` GROUP BY playlist_id
PlaylistSong Load (0.2ms) SELECT `playlist_songs`.* FROM `playlist_songs` WHERE `playlist_songs`.`id` IN (1, 4, 7)
Song Load (0.2ms) SELECT `songs`.* FROM `songs` WHERE `songs`.`id` IN (1, 4, 7)
Playlist Load (0.2ms) SELECT `playlists`.* FROM `playlists` WHERE `playlists`.`id` IN (1, 2, 3)
Playlist: Dubstep, Song: Dubstep song 1
Playlist: Top Rated, Song: Top Rated song 1
Playlist: Last Played, Song: Last Played song 1
This solution has some benefits:
Limited to 4 select statements
Does not load all playlist_songs - aggregating on db side
Does not load all songs - filtering by id's on db side
Tested with MySQL.
This will not show empty playlists.
And there could be problems with some DBs when playlists count > 1000

Answer 6

just fetch the song from the other side :
Song
.joins( :playlist )
.where( playlists: {id: [1,2,3]} )
.first
however, as #Dave S. suggested, "first" song in a playlist is random unless you explicitly specify an order (positioncolumn, or anything else) because SQL does not warrant the order in which the records are returned, unless you explicitly ask it.
EDIT
Sorry, I misread your question. I think that indeed a position column is necessary.
Song
.joins( :playlist )
.where( playlists: {id: [1,2,3]}, songs: {position: 1} )
If you do not want any position column at all, you can always try to group the songs by playlist id, but you'll have to select("songs.*, playlist_songs.*"), and the "first" song is still random. Another option is to use the RANK window function, but it is not supported by all RDBMS (for all i know, postgres and sql server do).

Answer 7

you can create a has_one association which, in effect, will call the first song that is associated to the playlist
class PlayList < ActiveRecord::Base
has_one :playlist_cover, class_name: 'Song', foreign_key: :playlist_id
end
Then just use this association.
Playlist.joins(:playlist_cover)
UPDATE: didn't see the join table.
you can use a :through option for has_one if you have a join table
class PlayList < ActiveRecord::Base
has_one :playlist_song_cover
has_one :playlist_cover, through: :playlist_song_cover, source: :song
end

Answer 8

Playlyst.joins(:playlist_songs).group('playlists.name').minimum('songs.name').to_a
hope it works :)
got this :
Product.includes(:vendors).group('products.id').collect{|product| [product.title, product.vendors.first.name]}
Product Load (0.5ms) SELECT "products".* FROM "products" GROUP BY products.id
Brand Load (0.5ms) SELECT "brands".* FROM "brands" WHERE "brands"."product_id" IN (1, 2, 3)
Vendor Load (0.4ms) SELECT "vendors".* FROM "vendors" WHERE "vendors"."id" IN (2, 3, 1, 4)
=> [["Computer", "Dell"], ["Smartphone", "Apple"], ["Screen", "Apple"]]
2.0.0p0 :120 > Product.joins(:vendors).group('products.title').minimum('vendors.name').to_a
(0.6ms) SELECT MIN(vendors.name) AS minimum_vendors_name, products.title AS products_title FROM "products" INNER JOIN "brands" ON "brands"."product_id" = "products"."id" INNER JOIN "vendors" ON "vendors"."id" = "brands"."vendor_id" GROUP BY products.title
=> [["Computer", "Dell"], ["Screen", "Apple"], ["Smartphone", "Apple"]]

Answer 9

You could add activerecord scope to your models to optimize how the sql queries work for you in the context of the app. Also, scopes are composable, thus make it easier to obtain what you're looking for.
For example, in your Song model, you may want a first_song scope
class Song < ActiveRecord::Base
scope :first_song, order("id asc").limit(1)
end
And then you can do something like this
playlists.songs.first_song
Note, you may also need to add some scopes to your PlaylistSongs association model, or to your Playlist model.

Answer 10

You didn't say if you had timestamps in your database. If you do though, and your records on the join table PlaylistSongs are created when you add a song to a playlist, I think this may work:
first_song_ids = Playlist.joins(:playlist_songs).order('playlist_songs.created_at ASC').pluck(:song_id).uniq
playlist_ids = Playlist.joins(:playlist_songs).order('playlist_songs.created_at ASC').pluck(:playlist_id).uniq
playlist_names = Playlist.where(id: playlist_ids).pluck(:playlist_name)
song_names = Song.where(id: first_song_ids).pluck(:song_name)
I believe playlist_names and song_names are now mapped by their index in this way. As in: playlist_names[0] first song name is song_names[0], and playlist_names[1] first song name is song_names[1] and so on. I'm sure you could combine them in a hash or an array very easily with built in ruby methods.
I realize you were looking for an efficient way to do this, and you said in the comments you didn't want to use a block, and I am unsure if by efficient you meant an all-in-one query. I am just getting used to combining all these rails query methods and perhaps looking at what I have here, you can modify things to your needs and make them more efficient or condensed.
Hope this helps.