Question

我有UserType和User模型。该协会如下;

UserType
has_one :user

User
belongs_to user_type

我试图让用户的user_type名称为X.

我试过了;

User.joins(:user_type).where(:user_type => {name: 'X'}).all

它给出了错误。

 `require'
    from /app/bin/rails:8:in `<main>'irb(main):015:0> User.joins(:user_type).where(:user_type => {name: 'normal'}).all
PG::UndefinedTable: ERROR:  missing FROM-clause entry for table "user_type"
LINE 1: ...sers" ON "user_types"."id" = "users"."user_type_id" WHERE "user_type"."u...
                                                             ^
: SELECT "users".* FROM "users" INNER JOIN "user_types" ON "user_types"."id" = "users"."user_type_id" WHERE "user_type"."name" = $1
PG::UndefinedTable: ERROR:  missing FROM-clause entry for table "user_type"
LINE 1: ...sers" ON "user_types"."id" = "users"."user_type_id" WHERE "user_type"."u...
                                                             ^
: SELECT "users".* FROM "users" INNER JOIN "user_types" ON "user_types"."id" = "users"."user_type_id" WHERE "user_type"."name" = $1
ActiveRecord::StatementInvalid: PG::UndefinedTable: ERROR:  missing FROM-clause entry for table "user_type"
LINE 1: ...sers" ON "user_types"."id" = "users"."user_type_id" WHERE "user_type"."u...

Answer 1

我不确定您的关联设置是否正确

您需要在user_type

之前添加冒号

User
belongs_to :user_type

您应该在联接查询中使用table name。

默认情况下，user_type的表名为user_types

User.joins(:user_type).where(:user_types => {name: 'X'}).all

如果你想避免n + 1问题，你应该使用includes

User.includes(:user_type).where(:user_types => {name: 'X'}).all

有一个属于连接

1 个答案: