在我的项目中,我使用 lombok 来避免为类编写 getter 和 setter。 此外,我正在使用 lombok.Builder 构建一个对象,而不是编写 new Object() 然后设置所有值。
但是当我们有继承关系时,当我们想使用 lombok builder 构造子对象时,我得不到父字段。
例如:
@Data
@NoArgsConstructor
@AllArgsConstructor
@ToString
@EqualsAndHashCode
public class Parent{
private String nationality;
.
.
// more columns
}
子类应该是这样的:
@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();
}
}
请告诉我,在 lombok 中有什么方法可以处理这种情况。
最佳答案
@Builder
无法确定您希望公开哪些 Parent
字段。
当 @Builder
放在类上时,只有在该类上显式声明的字段才会添加到 *Builder
。
当 @Builder
放在静态方法或构造函数上时,生成的 *Builder
将为每个参数提供一个方法。
此外,如果您使用的是 @Builder
,那么是否可以安全地假设至少 Child
是不可变的?
我提供了两个例子,一个是 Parent
是可变的而 Child
是不可变的,另一个是 Parent
和 Child
code> 是不可变的。
不可变父子
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());
}
}
可变父级,不可变子级:
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());
}
}
关于java - 当我们使用 lombok builder 有继承关系时,如何构建一个对象?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32989562/