Question

所以说我有一组看起来像这样的用户：

[19] pry(main)> j.connected_users
   (1.2ms)  SELECT "memberships"."invited_id" FROM "memberships"  WHERE (memberships.user_id = 117 OR memberships.invited_id = 117) AND ("memberships"."invited_id" IS NOT NULL) AND ("memberships"."invited_id" != 117)
  User Load (0.6ms)  SELECT "users".* FROM "users"  WHERE "users"."id" IN (1, 3)
=> [#<User id: 3, email: "jackie@test.com", encrypted_password: "$2a$10$kZWVB9Vu.08Dyesc74w20eod4Yh4sBnQ1XE1C2P9fXH...", reset_password_token: nil, reset_password_sent_at: nil, remember_created_at: nil, sign_in_count: 10, current_sign_in_at: "2016-02-21 08:21:13", last_sign_in_at: "2016-01-12 19:08:08", current_sign_in_ip: "127.0.0.1", last_sign_in_ip: "127.0.0.1", created_at: "2015-07-25 02:20:19", updated_at: "2016-02-21 08:21:13", first_name: "Jackie", confirmation_token: nil, confirmed_at: "2015-07-25 02:20:20", confirmation_sent_at: "2015-07-25 02:20:19", unconfirmed_email: nil, invitation_relation: nil, avatar: nil, invitation_token: nil, invitation_created_at: nil, invitation_sent_at: nil, invitation_accepted_at: nil, invitation_limit: nil, invited_by_id: nil, invited_by_type: nil, invitations_count: 0, bio: nil, last_name: "Gayle", gender: 1, birthday: nil, storage_used_in_bytes: 0>,
 #<User id: 1, email: "abc@test.com", encrypted_password: "$2a$10$IQXTVqZmMaUUv4WZlwXze.OTfWki2d2qZEjj01isCZZ...", reset_password_token: nil, reset_password_sent_at: nil, remember_created_at: nil, sign_in_count: 22, current_sign_in_at: "2016-03-09 01:06:39", last_sign_in_at: "2016-03-08 19:55:56", current_sign_in_ip: "127.0.0.1", last_sign_in_ip: "127.0.0.1", created_at: "2015-07-25 02:20:15", updated_at: "2016-03-09 01:06:39", first_name: "Marc", confirmation_token: nil, confirmed_at: "2015-07-25 02:20:16", confirmation_sent_at: "2015-07-25 02:20:16", unconfirmed_email: nil, invitation_relation: nil, avatar: "Marc-bio-pic.jpg", invitation_token: nil, invitation_created_at: nil, invitation_sent_at: nil, invitation_accepted_at: nil, invitation_limit: nil, invited_by_id: nil, invited_by_type: nil, invitations_count: 0, bio: "I am Marc Gayle. I build things and I love my fami...", last_name: "Gayle", gender: 0, birthday: nil, storage_used_in_bytes: 0>]

每个User has_many :nodes，因此当我查询每个nodes对象的user时，它会返回nodes的集合，如下所示：

[21] pry(main)> marc.id
=> 1
[22] pry(main)> marc.nodes
=> [#<Node id: 52, name: "High Knees", family_tree_id: 1, user_id: 1, media_id: 52, media_type: "Video", created_at: "2016-02-17 04:44:11", updated_at: "2016-03-08 22:21:52", circa: nil, is_comment: nil, cached_votes_total: 0, cached_votes_score: 0, cached_votes_up: 0, cached_votes_down: 0, cached_weighted_score: 0, cached_weighted_total: 0, cached_weighted_average: 0.0, cached_user_tag_list: "", cached_num_user_tags: 0, cached_tagged_user_names: [], comments_count: 0, description: "10PP Form Video High Knees">,
 #<Node id: 64, name: "Soaked", family_tree_id: 1, user_id: 1, media_id: 64, media_type: "Video", created_at: "2016-03-08 17:59:46", updated_at: "2016-03-08 22:21:54", circa: nil, is_comment: nil, cached_votes_total: 1, cached_votes_score: 1, cached_votes_up: 1, cached_votes_down: 0, cached_weighted_score: 1, cached_weighted_total: 1, cached_weighted_average: 0.0, cached_user_tag_list: "", cached_num_user_tags: 0, cached_tagged_user_names: [], comments_count: 0, description: "10PP Soaked Workout">]

如何根据以上集合获得全面的nodes集合，如果不手动迭代集合，则没有重复项？

我可以使用joins或其他一些ActiveRecord方法实现此目的吗？

Answer 1

您不需要关联：

Node.where(user_id: j.connected_users.map(&:id))

如果我有一个集合，如何在不使用每个项目的情况下获取集合中每个项目的所有关联对象？

1 个答案: