json - Spring Boot json 在没有默认构造函数的情况下建模

Question

我会遇到一个问题，但我无法解决。

上下文

我有一个 Controller :

@RestController
@RequestMapping("/classroom")
public class ClassroomController {

    @Inject
    private ClassroomManager classroomManager;

    @RequestMapping(value= "", method = RequestMethod.POST)
    public Classroom create(Classroom classroom) {
        ... some checks ...
        return classroomManager.create(classroom);
    }
}

模型类:

public class Classroom {

    private Long id;
    private final String name;
    private final RoomCapacity roomCapacity;
    private final Set<RoomEquipment> equipments = new HashSet<>();
    [... No setter, No default constructor ...]



    public Classroom(String name, RoomCapacity roomCapacity) {
        [... some checks ...]
        this.name = name;
        this.roomCapacity = roomCapacity;
    }
}

问题是，我无法以 JSON 格式发送带有 Classroom 的 POST 请求。

当我这样做时，我会遇到这个异常:

org.springframework.web.util.NestedServletException: Request processing failed; nested exception is org.springframework.beans.BeanInstantiationException: Failed to instantiate [org.ubp.ent.backend.core.model.classroom.Classroom]: No default constructor found; nested exception is java.lang.NoSuchMethodException: org.ubp.ent.backend.core.model.classroom.Classroom.<init>()

我的问题是什么

我确实理解错误，我可能会向类添加一个默认构造函数。
但是点 Classroom 来自模型，我不想允许默认实例化。

如何让 Spring 能够从 Json 构建？

Answer 1

您可以通过为您的模型类定义 Jackson 自定义反序列化器来实现。查看下面的链接，了解如何实现和配置它。

http://wiki.fasterxml.com/JacksonHowToCustomDeserializers

http://www.baeldung.com/jackson-deserialization