如何在单个查询中获得此查询结果而不是N + 1

Question

A pool_tournament有许多pool_tournament_matches，每个匹配属于多个users。用户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通过协会在这里。一个是在user和pool_tournament之间。另一个是在pool_tournament_match和user之间。

我的查询是找出哪个pool_tournament_matches只有1个用户。我的查询向我提供了匹配列表，但它为每个pool_tournament_match执行N + 1查询。

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，可以张贴架构。

谢谢！

Answer 1

您可以让SQL为您计数。下面以Postgres的应该工作（不知道其他数据库）：

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 } 

一切行动和incuding的.group产生一个单一的查询，并将它附加一个“USER_COUNT”属性则返回pool_tournament_matches。因此，最后的.select发生在内存中，对结果进行解析而不进行额外的数据库调用。