如何在單個查詢中獲得此查詢結果而不是N + 1

Question

A pool_tournament有許多pool_tournament_matches，每個匹配屬於多個users。用戶has_many pool_tournaments和has_many pool_tournament_matches。

pool_tournament.rb

has_many :pool_tournament_matches 

pool_tournament_match.rb

belongs_to :pool_tournament 
has_many :pool_tournament_match_users, class_name: 'PoolTournamentMatchUser' 
has_many :users, through: :pool_tournament_match_users 

user.rb

has_many :pool_tournament_users, class_name: 'PoolTournamentUser' 
has_many :pool_tournaments, through: :pool_tournament_users 

has_many :pool_tournament_match_users, class_name: 'PoolTournamentMatchUser' 
has_many :pool_tournament_matches, through: :pool_tournament_match_users 

有2的has_many通過協會在這裏。一個是在user和pool_tournament之間。另一個是在pool_tournament_match和user之間。

我的查詢是找出哪個pool_tournament_matches只有1個用戶。我的查詢向我提供了匹配列表，但它爲每個pool_tournament_match執行N + 1查詢。

tournament.pool_tournament_matches.includes(:users).select { |m| m.users.count == 1 } 

PoolTournamentMatch Load (0.6ms) SELECT "pool_tournament_matches".* FROM "pool_tournament_matches" WHERE "pool_tournament_matches"."pool_tournament_id" = $1 [["pool_tournament_id", 2]] PoolTournamentMatchUser Load (0.6ms) SELECT "pool_tournament_match_users".* FROM "pool_tournament_match_users" WHERE "pool_tournament_match_users"."pool_tournament_match_id" IN (1, 2, 3, 4) User Load (0.6ms) SELECT "users".* FROM "users" WHERE "users"."id" IN (1, 2, 3, 4, 5, 6, 7, 8) (0.8ms) SELECT COUNT(*) FROM "users" INNER JOIN "pool_tournament_match_users" ON "users"."id" = "pool_tournament_match_users"."user_id" WHERE "pool_tournament_match_users"."pool_tournament_match_id" = $1 [["pool_tournament_match_id", 1]] (0.7ms) SELECT COUNT(*) FROM "users" INNER JOIN "pool_tournament_match_users" ON "users"."id" = "pool_tournament_match_users"."user_id" WHERE "pool_tournament_match_users"."pool_tournament_match_id" = $1 [["pool_tournament_match_id", 2]] (0.7ms) SELECT COUNT(*) FROM "users" INNER JOIN "pool_tournament_match_users" ON "users"."id" = "pool_tournament_match_users"."user_id" WHERE "pool_tournament_match_users"."pool_tournament_match_id" = $1 [["pool_tournament_match_id", 3]] (0.7ms) SELECT COUNT(*) FROM "users" INNER JOIN "pool_tournament_match_users" ON "users"."id" = "pool_tournament_match_users"."user_id" WHERE "pool_tournament_match_users"."pool_tournament_match_id" = $1 [["pool_tournament_match_id", 4]]

我也不介意使用RAW SQL，如果需要，可以張貼架構。

謝謝！

Answer 1

您可以讓SQL爲您計數。下面以Postgres的應該工作（不知道其他數據庫）：

tournament.pool_tournament_matches 
    .select("pool_tournament_matches.*, COUNT(users.id) as user_count") 
    .joins("LEFT OUTER JOIN pool_tournament_match_users ON (pool_tournament_match_users.pool_tournament_match_id = pool_tournament_matches.id)") 
    .joins("LEFT OUTER JOIN users ON (pool_tournament_match_users.user_id = users.id)") 
    .group("pool_tournament_matches.id") 
    .select { |match| match.user_count > 0 } 

一切行動和incuding的.group產生一個單一的查詢，並將它附加一個「USER_COUNT」屬性則返回pool_tournament_matches。因此，最後的.select發生在內存中，對結果進行解析而不進行額外的數據庫調用。