比较 JPA 实体和 DTO

Question

我需要一种标准方法来将 JPA 实体与其 DTO 进行比较，并确定它们是否代表相同的业务对象。我可以想到三种方法，每个 DTO 上的自定义方法，带有静态方法或比较器的接口。

基于 João Dias 的答案，方法 4 - 继承。

Pros/Cons

方法 1 - 一直都不好
方法 2 - 使用接口支持组合而不是继承，但需要使用自定义方法名称的语义 (businessKeysMatch())
方法 3 - 不需要修改源代码
方法 4 - 简化语义，因为它使用标准 equals() 但需要“样板 equals()/hashcode()

任何其他 pros/cons 方法或对其他方法的建议？

最终，我选择了方法二（Interface）。继承方法显示出希望，但作为 JPA 实体的 classes 之一使映射比我想要的更复杂。

感谢阅读和思考我的问题！

背景

实体

一个数据库键
在确定相等性的 ORM 和数据库中强制执行的唯一业务密钥 (equals()/hashcode())
public 属性（姓名、地址、年龄等）

非public/机密属性（密码、刷新令牌、SSN 等）

  @Data
  @EqualsAndHashCode(onlyExplicitlyIncluded = true)
  @Entity
  @Table(uniqueConstraints = { @UniqueConstraint(columnNames = { "businessKey1", "businessKey2" }) })
  class UserEntity {
      @Id
      Long id;
      @NotNull
      @EqualsAndHashCode.Include
      Long businessKey1;
      @NotNull
      @EqualsAndHashCode.Include
      Long businessKey2;
      String name;
      Integer age;
      String password;
      String refreshToken;
      String SSN;
  }

DTO（完整）

确定相等性的业务键(equals()/hashcode())

public 属性（姓名、地址、年龄等）

  @Data
  @EqualsAndHashCode(onlyExplicitlyIncluded = true)
  class UserDto {
      @EqualsAndHashCode.Include
      @NotNull
      Long businessKey1;
      @EqualsAndHashCode.Include
      @NotNull
      Long businessKey2;
      String name;
      String address;
      Integer age;
  }

DTO（有限）

确定相等性的业务键(equals()/hashcode())

已选择public个属性（名称）

  @Data
  @EqualsAndHashCode(onlyExplicitlyIncluded = true)
  class UserNameDto {
      @EqualsAndHashCode.Include
      @NotNull
      Long businessKey1;
      @EqualsAndHashCode.Include
      @NotNull
      Long businessKey2;
      String name;
  }

方法 1 - 添加到每个 User*Dto

的自定义方法

    boolean businessKeysMatch(UserEntity entity) {
        if((this.getBusinessKey1() == entity.getBusinessKey1()) && (this.getBusinessKey2() == entity.getBusinessKey2()))
            return true;
        return false;
    }

方法 2 - 将静态方法添加到公共接口

    interface UserKeys {
        Long getBusinessKey1();
        Long getBusinessKey2();
        static boolean businessKeysMatch(UserKeys o1, UserKeys o2) {
            if((o1.getBusinessKey1() == o2.getBusinessKey1()) && (o1.getBusinessKey2() == o2.getBusinessKey2()))
                return true;
            return false;
        }
    }

    class UserEntity implements UserKeys {
        // no other changes
    }

    class UserDto implements UserKeys {
        // no other changes
    }

    class UserEntity implements UserKeys {
        // no other changes
    }

方法 3 - 比较器

    interface UserBusinessKey {
        Long getBusinessKey1();
        Long getBusinessKey2();
    }

    class UserDto implements UserCompare {
        // no other changes
    }

    class UserEntity implements UserCompare {
        // no other changes
    }

    class UserCompare implements Comparator<UserBusinessKey> {
        public int compare(UserBusinessKey o1, UserBusinessKey o2) {
            int key1Compare = o1.getBusinessKey1().compareTo(o2.getBusinessKey1());

            if (key1Compare == 0) 
                return o1.getBusinessKey2().compareTo(o2.getBusinessKey2());
            return key1Compare;
        }
    }

方法 4 - 仅继承基数 class 的 equals/hashcode

    @SuperBuilder
    @AllArgsConstructor
    @NoArgsConstructor
    @Data
    abstract class UserBase {
        @NotNull
        Long businessKey1;
        @NotNull
        Long businessKey2;
        // lombok generates a standard equals() / hashcode() pair
    }

    @SuperBuilder
    @Getter
    @Setter
    @ToString
    @Entity
    @Table(uniqueConstraints = { @UniqueConstraint(columnNames = { "businessKey1", "businessKey2" }) })
    class UserEntity extends UserBase {
        @Id
        Long id;
        String name;
        Integer age;
        String password;
        String refreshToken;
        String SSN;
        
        // handcoded equals/hashcode that only call super
        @Override
        public boolean equals(Object obj) {
            return super.equals(obj);
        }

        @Override
        public int hashCode() {
            return super.hashCode();
        }
    }

    @SuperBuilder
    @Getter
    @Setter
    @ToString
    class UserDto extends UserBase {
        String name;
        String address;
        Integer age;
        
        // handcoded equals/hashcode that only call super
        @Override
        public boolean equals(Object obj) {
            return super.equals(obj);
        }

        @Override
        public int hashCode() {
            return super.hashCode();
        }
    }

Answer 1

如果您有多个 User*Dto，我会创建一个抽象 AbstractUserDto，然后由所有具体的用户 DTO 对其进行扩展。您可以在那里放置您在方法 1 中显示的方法（这样您就不会一遍又一遍地复制相同的代码）：

@Data
@EqualsAndHashCode(onlyExplicitlyIncluded = true)
public abstract class AbstractUserDto {
    @EqualsAndHashCode.Include
    @NotNull
    Long businessKey1;
    
    @EqualsAndHashCode.Include
    @NotNull
    Long businessKey2;
    
    String name;

    public final boolean businessKeysMatch() {
        return (this.getBusinessKey1() == entity.getBusinessKey1()) && (this.getBusinessKey2() == entity.getBusinessKey2());
    }
}

比较 JPA 实体和 DTO

Compare JPA Entity and DTO

java

jpa

lombok

spring-data-jpa