ruby-on-rails – 如何为组合字段的唯一性建模此复杂验证
链接有两个组件:componenta_id和componentb_id.为此,在Link模型文件中我有:
belongs_to :componenta,class_name: "Component" belongs_to :componentb,class_name: "Component" validates :componenta_id,presence: true validates :componentb_id,presence: true validates :componenta_id,uniqueness: { scope: :componentb_id } validates :componentb_id,uniqueness: { scope: :componenta_id } 并在迁移文件中: create_table :links do |t| t.integer :componenta_id,null: false t.integer :componentb_id,null: false ... end add_index :links,:componenta_id add_index :links,:componentb_id add_index :links,[:componenta_id,:componentb_id],unique: true 问题:这一切都有效.现在我希望componanta和componentb的组合是独一无二的,无论它们的顺序如何.因此,无论哪个组件是组件a,哪个组件是组件b(毕竟这是相同的链接;两个相同组件之间的链接).因此,不应允许下面的两个记录,因为它们代表相同的链接,因此不是唯一的: > componenta_id = 1; componentb_id = 2 如何创建此唯一性验证?我有模型验证工作(见下文),但想知道是否以及如何在迁移/数据库级别添加验证…? 模型验证 before_save :order_links validates :componenta_id,uniqueness: { scope: :componentb_id } private def order_links if componenta_id > componentb_id compb = componentb_id compa = componenta_id self.componenta_id = compb self.componentb_id = compa end end 以下测试证实了上述工作: 1. test "combination of two links should be unique" do 2. assert @link1.valid? 3. assert @link2.valid? 4. @link1.componenta_id = 3 #@link2 already has combination 3-4 5. @link1.componentb_id = 4 6. assert_not @link1.valid? 7. @link1.componenta_id = 4 8. @link1.componentb_id = 3 9. assert_raises ActiveRecord::RecordNotUnique do 10. @link1.save 11. end 12.end 迁移/数据库验证: 解决方法
也许可以通过以下方式控制链接的创建:
def create_unique_link( comp_1,comp_2 ) if comp_1.id > comp_2.id first_component = comp_1 second_component = comp_2 end link = Link.find_or_create_by( componenta_id: first_comp.id,componentb_id: second_comp.id ) end 如果您需要验证,那么您可以自定义验证: def ensure_uniqueness_of_link if comp_1.id > comp_2.id first_component = comp_1 second_component = comp_2 end if Link.where( componenta_id: first_component.id,componentb_id: second_component ).first errors.add( :link,'Links should be unique' ) end end (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |