当我们使用lombok builder有继承关系时如何构建对象？

Question

在我的项目中,我使用lombok来避免为一个类编写getter和setter.此外,我使用lombok.Builder来构建一个对象,而不是编写新的Obeject(),然后设置所有的值.

但是当我们有继承关系并且当我们想要使用lombok builder构建子对象时,我没有得到父类的字段.

例如:

@Data
@NoArgsConstructor
@AllArgsConstructor
@ToString
@EqualsAndHashCode
public class Parent{
  private String nationality;
  .
  .
  // more columns
}

而Child类将是这样的:

@Data
@Builder
@NoArgsConstructor
@AllArgsConstructor
@ToString(callSuper = true)
@EqualsAndHashCode(callSuper = true)
public class Child extends Parent{
   private String firstName;
   private String lastName;
   .
   .
}

在我的测试类中,我需要构建子对象

public class Test{

 public void testMethod(){
   Child child = Child.builder()
            .firstName("Rakesh")
            .lastName("SS")
            .nationality("some text")// I am not able to set nationality               
            .build();
 }


}

请让我知道,有没有办法在lombok中处理这种情况.

Answer 1

@Builder无法确定Parent您希望公开哪些字段.

当@Builder放置在类上时,只有在该类上显式声明的字段才会添加到该类中*Builder.

当@Builder放置在静态方法或构造函数上时,结果*Builder将为每个参数提供一个方法.

此外,如果你正在使用@Builder那么可以安全地假设至少Child是不可变的吗？

我公司供应的两个例子,其中一个Parent是可变的并且Child是不变的,一个其中两个Parent和Child是不可变的.

不可改变的父母和子女

import static org.junit.Assert.*;
import lombok.Builder;
import lombok.EqualsAndHashCode;
import lombok.ToString;
import lombok.Value;
import lombok.experimental.NonFinal;

import org.junit.Test;

public class So32989562ValueTest {

    @Value
    @NonFinal
    public static class Parent {

        protected final String nationality;

    }

    @Value
    @ToString(callSuper = true)
    @EqualsAndHashCode(callSuper = true)
    public static class Child extends Parent {

        private final String firstName;

        private final String lastName;

        @Builder(toBuilder = true)
        private Child(String nationality, String firstName, String lastName) {
            super(nationality);
            this.firstName = firstName;
            this.lastName = lastName;
        }
    }

    @Test
    public void testChildBuilder() {

        String expectedFirstName = "Jeff";
        String expectedLastName = "Maxwell";
        String expectedNationality = "USA";

        Child result = Child.builder()
            .firstName(expectedFirstName)
            .lastName(expectedLastName)
            .nationality(expectedNationality)
            .build();

        assertEquals(result.toString(), expectedFirstName, result.getFirstName());
        assertEquals(result.toString(), expectedLastName, result.getLastName());
        assertEquals(result.toString(), expectedNationality, result.getNationality());
    }
}

可变父母,不可变子女:

import static org.junit.Assert.*;
import lombok.Builder;
import lombok.Data;
import lombok.EqualsAndHashCode;
import lombok.ToString;
import lombok.Value;

import org.junit.Test;

public class So32989562DataTest {

    @Data
    public static class Parent {

        protected String nationality;

    }

    @Value
    @ToString(callSuper = true)
    @EqualsAndHashCode(callSuper = true)
    public static class Child extends Parent {

        private final String firstName;

        private final String lastName;

        @Builder(toBuilder = true)
        private Child(String nationality, String firstName, String lastName) {
            this.setNationality(nationality);
            this.firstName = firstName;
            this.lastName = lastName;
        }
    }

    @Test
    public void testChildBuilder() {

        String expectedFirstName = "Jeff";
        String expectedLastName = "Maxwell";
        String expectedNationality = "USA";

        Child result = Child.builder()
            .firstName(expectedFirstName)
            .lastName(expectedLastName)
            .nationality(expectedNationality)
            .build();

        assertEquals(result.toString(), expectedFirstName, result.getFirstName());
        assertEquals(result.toString(), expectedLastName, result.getLastName());
        assertEquals(result.toString(), expectedNationality, result.getNationality());
    }
}

Answer 2

以上解决方案有效,但需要太多的解决方法.而且,子类和父类的任何更改都需要在任何地方更改构造函数参数.

Lombok针对Builder注释面临的继承问题引入了版本:1.18.2的实验性功能,可以使用@SuperBuilder注释解决如下.

@SuperBuilder
public class ParentClass {
    private final String a;
    private final String b;
}

@SuperBuilder
public class ChildClass extends ParentClass{
    private final String c;
}

现在,可以使用如下的Builder类(使用@Builder注释无法实现)

ChildClass.builder().a("testA").b("testB").c("testC").build();