首先是我的Gemfile:
- source 'https://rubygems.org'
- gem 'rails','~> 4.1.6'
- gem 'pg'
下一个.我的模特:
- class User < ActiveRecord::Base
- end
- class Teacher < User
- has_and_belongs_to_many :resources,foreign_key: :user_id
- end
- class Resource < ActiveRecord::Base
- has_and_belongs_to_many :teachers,association_foreign_key: :user_id
- end
原始数据库数据:
- select * from resources;
- id | created_at | updated_at
- ----+----------------------------+----------------------------
- 1 | 2014-10-13 08:24:07.308361 | 2014-10-13 08:24:07.308361
- 2 | 2014-10-13 08:24:07.889907 | 2014-10-13 08:24:08.156898
- 3 | 2014-10-13 08:24:08.68579 | 2014-10-13 08:24:08.884731
- 4 | 2014-10-13 08:24:09.997244 | 2014-10-13 08:24:10.205753
- (4 rows)
- select * from users;
- id | created_at | updated_at | type
- ----+----------------------------+----------------------------+---------
- 13 | 2014-10-13 08:24:01.086192 | 2014-10-13 08:24:01.086192 | Teacher
- 12 | 2014-10-13 08:24:00.984957 | 2014-10-13 08:24:00.984957 | Teacher
- 2 | 2014-10-13 08:23:59.950349 | 2014-10-16 08:46:02.531245 | Teacher
- (3 rows)
- select * from resources_users;
- user_id | resource_id
- ---------+-------------
- 13 | 1
- 2 | 2
- 12 | 3
- 2 | 4
- (4 rows)
最后这个bug:
- ➜ rails_test bundle exec rails c
- Loading development environment (Rails 4.1.6)
- 2.1.2 :001 > Resource.all.includes(:teachers).map(&:teachers).map(&:to_a)
- Resource Load (0.6ms) SELECT "resources".* FROM "resources"
- sql (1.3ms) SELECT "resources_users".*,"resources_users"."user_id" AS t0_r0,"resources_users"."resource_id" AS t0_r1,"users"."id" AS t1_r0,"users"."created_at" AS t1_r1,"users"."updated_at" AS t1_r2,"users"."type" AS t1_r3 FROM "resources_users" LEFT OUTER JOIN "users" ON "users"."id" = "resources_users"."user_id" AND "users"."type" IN ('Teacher') WHERE "users"."type" IN ('Teacher') AND "resources_users"."resource_id" IN (1,2,3,4)
- => [
- [#<Teacher id: 13,created_at: "2014-10-13 08:24:01",updated_at: "2014-10-13 08:24:01",type: "Teacher">],[],[]]
正如您所看到的,只有第一批教师返回收藏中.但是Rails生成的sql是正确的并返回所有数据:
- SELECT "resources_users".*,4);
- user_id | resource_id | t0_r0 | t0_r1 | t1_r0 | t1_r1 | t1_r2 | t1_r3
- ---------+-------------+-------+-------+-------+----------------------------+----------------------------+---------
- 13 | 1 | 13 | 1 | 13 | 2014-10-13 08:24:01.086192 | 2014-10-13 08:24:01.086192 | Teacher
- 2 | 2 | 2 | 2 | 2 | 2014-10-13 08:23:59.950349 | 2014-10-16 08:46:02.531245 | Teacher
- 12 | 3 | 12 | 3 | 12 | 2014-10-13 08:24:00.984957 | 2014-10-13 08:24:00.984957 | Teacher
- 2 | 4 | 2 | 4 | 2 | 2014-10-13 08:23:59.950349 | 2014-10-16 08:46:02.531245 | Teacher
- (4 rows)
以前有人遇到过这样的问题吗?我无法理解这里发生了什么.
附:如果你做Resource.all.includes(:教师).map {| r | r.reload.teachers}结果是正确的.然而,它从包含中删除了感觉并提供了N 1问题.
更新:还有一个值得一提的发现.如果我删除STI一切正常.
解决方法
IRB(主):017:0> Resource.all.includes(:教师).MAP(安培;:教师).MAP(安培;:to_a)资源负载(0.6ms)选择“资源”.* FROM“资源”sql(6.9ms)SELECT“resources_users”.*,“resources_users”.“id”AS t0_r0,“resources_users”.“resource_id”AS t0_r1,“resources_users”.“user_id”AS t0_r2,“users”.“id”AS t1_r0,“users”.“type”AS t1_r1,“users”.“created_at”AS t1_r2,“users”.“updated_at”AS t1_r3 FROM“resources_users”LEFT OUTER JOIN“users”ON“users”.“id”= “resources_users”.“user_id”和“users”.“type”IN(‘Teacher’)WHERE“users”.“type”IN(‘Teacher’)AND“resources_users”.“resource_id”IN(1,4)=> [[#<老师ID:13,输入:“老师”,created_at:“2015-11-05 07:02:59”,updated_at:“2015-11-05 07:02:59”>],[ #<教师ID:2,输入:“Teacher”,created_at:“2015-11-05 07:02:20”,updated_at:“2015-11-05 07:02:32”>],[#< ;教师ID:12,created_at:“2015-11-05 07:03:50”,updated_at:“2015-11-05 07:03:50”>],[#<老师id:2,updated_at:“2015-11-05 07:02:32”>]]