MapStruct：对象的映射列表，当对象从两个对象映射时

Question

假设我有这样的映射：

@Mapping(source = "parentId", target = "parent.id")
Child map(ChildDto dto, Parent parent);

现在我需要将 ChildDto 列表映射到 Child 列表，但它们都具有相同的父级。我希望做这样的事情：

List<Child> map(List<ChildDto> dtoList, Parent parent);

但它不起作用。有机会做到吗？

Answer 1

我找到了如何用装饰器实现它，谢谢@Gunnar 这是一个实现：

豆子

public class Child {
    int id;
    String name;
}
public class Parent {
    int id;
    String name;
}
public class ChildDto {
    int id;
    String name;
    int parentId;
    String parentName;
}
// getters/settes ommited

映射器

@Mapper
@DecoratedWith(ChildMapperDecorator.class)
public abstract class ChildMapper {
    public static final ChildMapper INSTANCE = Mappers.getMapper(ChildMapper.class);

    @Mappings({
            @Mapping(target = "parentId", ignore = true),
            @Mapping(target = "parentName", ignore = true)
    })
    @Named("toDto")
    abstract ChildDto map(Child child);

    @Mappings({
            @Mapping(target = "id", ignore = true),
            @Mapping(target = "name", ignore = true),
            @Mapping(target = "parentId", source = "id"),
            @Mapping(target = "parentName", source = "name")
    })
    abstract ChildDto map(@MappingTarget ChildDto dto, Parent parent);

    @IterableMapping(qualifiedByName = "toDto") // won't work without it
    abstract List<ChildDto> map(List<Child> children);

    List<ChildDto> map(List<Child> children, Parent parent) {
        throw new UnsupportedOperationException("Not implemented");
    }
}

装饰者

public abstract class ChildMapperDecorator extends ChildMapper {
    private final ChildMapper delegate;

    protected ChildMapperDecorator(ChildMapper delegate) {
        this.delegate = delegate;
    }

    @Override
    public List<ChildDto> map(List<Child> children, Parent parent) {
        List<ChildDto> dtoList = delegate.map(children);
        for (ChildDto childDto : dtoList) {
            delegate.map(childDto, parent);
        }
        return dtoList;
    }
}

我使用abstract class, 不interface用于映射器，因为如果interface您无法排除生成方法map(List<Child> children, Parent parent)，并且生成的代码在编译时无效。

Answer 2

从目前的情况来看，这是不可能的。您可以使用装饰器或后映射方法将父对象设置为所有子对象。