Rails PG :: UndefinedTable:错误:缺少表的FROM子句条目

我有模特

class Offer < ActiveRecord::Base belongs_to :agency end class Agency < ActiveRecord::Base has_many :offers end 

当我提出这样的要求时 – 一切都很好

 @offers = Offer.with_state(:confirmed). includes(:destination, :cruise_line, :ship). paginate(per_page: 10, page: params[:page]).decorate 

但我想只选择属于active代理商的商品( agencies表中的列state ),所以我尝试这样做:

 @offers = Offer.with_state(:confirmed). includes(:destination, :cruise_line, :ship). joins(:agency). where(agency: {state: 'active'}). paginate(per_page: 10, page: params[:page]).decorate 

完成此操作后,我收到错误PG::UndefinedTable: ERROR: missing FROM-clause entry for table "agency" 。 我的代码出了什么问题?

查询给了我这个错误和sql:

 PG::UndefinedTable: ERROR: missing FROM-clause entry for table "agency" LINE 1: ...id" WHERE ("offers"."state" IN ('confirmed')) AND "agency"."... ^ : SELECT "offers"."id" AS t0_r0, "offers"."name" AS t0_r1, "offers"."destination_id" AS t0_r2, "offers"."cruise_line_id" AS t0_r3, "offers"."ship_id" AS t0_r4, "offers"."departure_date" AS t0_r5, "offers"."departure_port_id" AS t0_r6, "offers"."arrival_date" AS t0_r7, "offers"."arrival_port_id" AS t0_r8, "offers"."flight_price" AS t0_r9, "offers"."bonus" AS t0_r10, "offers"."itinerary" AS t0_r11, "offers"."board_language_id" AS t0_r12, "offers"."agency_landing_page" AS t0_r13, "offers"."benefits" AS t0_r14, "offers"."inner_price" AS t0_r15, "offers"."inner_price_normal" AS t0_r16, "offers"."outer_price" AS t0_r17, "offers"."outer_price_normal" AS t0_r18, "offers"."balcony_price" AS t0_r19, "offers"."balcony_price_normal" AS t0_r20, "offers"."suite_price" AS t0_r21, "offers"."suite_price_normal" AS t0_r22, "offers"."lucky_price" AS t0_r23, "offers"."lucky_price_normal" AS t0_r24, "offers"."valid_from" AS t0_r25, "offers"."valid_till" AS t0_r26, "offers"."created_at" AS t0_r27, "offers"."updated_at" AS t0_r28, "offers"."description" AS t0_r29, "offers"."agency_id" AS t0_r30, "offers"."state" AS t0_r31, "destinations"."id" AS t1_r0, "destinations"."name" AS t1_r1, "destinations"."created_at" AS t1_r2, "destinations"."updated_at" AS t1_r3, "cruise_lines"."id" AS t2_r0, "cruise_lines"."name" AS t2_r1, "cruise_lines"."created_at" AS t2_r2, "cruise_lines"."updated_at" AS t2_r3, "ships"."id" AS t3_r0, "ships"."name" AS t3_r1, "ships"."picture" AS t3_r2, "ships"."cruise_line_id" AS t3_r3, "ships"."created_at" AS t3_r4, "ships"."updated_at" AS t3_r5 FROM "offers" INNER JOIN "agencies" ON "agencies"."id" = "offers"."agency_id" LEFT OUTER JOIN "destinations" ON "destinations"."id" = "offers"."destination_id" LEFT OUTER JOIN "cruise_lines" ON "cruise_lines"."id" = "offers"."cruise_line_id" LEFT OUTER JOIN "ships" ON "ships"."id" = "offers"."ship_id" WHERE ("offers"."state" IN ('confirmed')) AND "agency"."state" = 'active' LIMIT 10 OFFSET 0 

您添加的唯一内容是where -clause。 看起来像是有问题的:

 where(agency: {state: 'active'}) 

…并且产生错误的条件,因为在这种情况下,散列键应该表示表名。 你应该这样:

 where(agencies: {state: 'active'}) 

更新

看到这一点引起了很多关注,我想我也应该提出一种不同的方式来做同样的事情,稍微更具组合性:

 merge( Agency.where(state: 'active') ) 

这个怎么样更好? 它没有假设模型的表名(不是大多数时候都很重要),并允许您使用范围

 # Inside Agency scope :active, -> { where(state: 'active') } # Somewhere else merge(Agency.active)