查找其关联依次为零关联的所有记录。

Question

鉴于以下模式：

class Score < ApplicationRecord
  belongs_to :composition
end

class Composition < ApplicationRecord
  has_many :scores
  has_one :invoice, dependent: :destroy
end

class Invoice < ApplicationRecord
  belongs_to :composition
end

找到分数的最佳方法是什么，它的组成有一张发票

我试过：

Score.joins(:composition).where(composition: {invoice: nil})

csn = Composition.includes(:invoice).where(invoices:{id:nil})
Score.where(csn.include? composition)
Score.where(csn.map(&:id).include? composition_id)

Score.where(Composition.left_outer_joins(:invoice).where(invoices:{id:nil}).includes? composition)

都有错误。有什么想法吗？

编辑：这里是与schema.rb对应的表

  create_table "compositions", force: :cascade do |t|
    ...
  end

  create_table "invoices", force: :cascade do |t|
    t.integer  "composition_id"
    ...
    t.index ["composition_id"], name: "index_invoices_on_composition_id", using: :btree
  end

  create_table "scores", force: :cascade do |t|
    t.integer  "composition_id",                           null: false
    ...
  end

Answer 1

请尝试以下查询：

Score.joins(:composition).includes(composition: : invoice).where(invoices: { id: nil })

Answer 2

试着跟随

Score.joins(:composition).where('compositions.invoice_id IS NULL')

上面应该管用，祝你好运！

Answer 3

请尝试以下查询，

Score.joins(:composition).where('compositions.id NOT IN (?)', Invoice.pluck(:composition_id))

如果这不起作用，那么让我知道compositions表和scores表中的列