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Add check if there is a tie in scrobble count

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mitteneer 2025-07-02 16:10:55 -04:00
parent 5739f89e0d
commit 9fa90ff129
2 changed files with 15 additions and 8 deletions
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19
src/queries.zig
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@ -346,14 +346,17 @@ pub fn loadQuery(entity: EntityType, query_type: QueryTypeEnum) GeneratedQuery {
.entity_info => switch (entity) {
.scrobble => unreachable,
.song =>
\\SELECT * FROM
\\(SELECT *, TO_CHAR(ROW_NUMBER() OVER (ORDER BY scrobbles DESC),'FM99999th') AS rank FROM
\\(SELECT songs.name AS song_name, songs.id AS song_id, COUNT(scrobbles) AS scrobbles
\\FROM albumsongs
\\INNER JOIN songs ON albumsongs.song_id = songs.id
\\INNER JOIN scrobbles ON scrobbles.albumsong = albumsongs.id
\\GROUP BY songs.id))
\\WHERE song_id = $1
\\WITH ranked AS (
\\SELECT songs.name AS song_name, COUNT(songs.id) AS scrobbles, RANK() OVER ( ORDER BY COUNT(songs.id) DESC) AS rank, songs.id AS song_id
\\FROM scrobbles
\\INNER JOIN albumsongs ON albumsongs.id = scrobbles.albumsong
\\INNER JOIN songs ON songs.id = albumsongs.song_id
\\GROUP BY songs.id)
\\SELECT * FROM (SELECT song_name, scrobbles, TO_CHAR(rank,'FM9999th') AS rank, song_id,
\\( rank = lag(rank, 1, -1::bigint) OVER (ORDER BY rank)
\\OR rank = lead(rank, 1, -1::bigint) OVER (ORDER BY rank)
\\)AS is_tie
\\FROM ranked) WHERE song_id = $1;
,
.album =>
\\SELECT album_name, t.album_id, artists.name AS artist_name, artists.id AS artist_id, song_num, scrobbles, rank FROM