如何处理 REST 中的@OneToMany 关系

Question

我正在设计一个小的REST 应用程序，它允许执行一些基本操作。到目前为止一切顺利，我关注@Entity，称为Client，需要与@Entity 一起持久化，称为Loan：

客户：

@Entity
@Table(name = "clients")
public class Client{

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "CLIENT_ID")
    private Long id;

    private String name;
    private String surname;
    private String email;
    private String phone;

    @OneToMany(fetch = FetchType.LAZY, mappedBy = "client")
    private Set<Loan> loans;

    public Client() {}

    public Client(Long id, String name, String surname, String email, String phone) {
        this.id = id;
        this.name = name;
        this.surname = surname;
        this.email = email;
        this.phone = phone;
    }

   // getters/setters

}

贷款：

@Entity
@Table(name = "loans")
public class Loan{

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "CLIENT_ID")
    private Client client;

    @Temporal(TemporalType.DATE)
    private Date startDate = new Date();

    @Temporal(TemporalType.DATE)
    private Date originTerm;

    private Float maxPossibleAmount;

    private String notes;

    public Loan() {}

    public Loan(Long id, Client client, Date startDate, Date originTerm, Float maxPossibleAmount, String notes) {
        this.id = id;
        this.client = client;
        this.startDate = startDate;
        this.originTerm = originTerm;
        this.maxPossibleAmount = maxPossibleAmount;
        this.notes = notes;
    }

 // getters/setters
}

注册客户可以通过邮递员完美运行，但是我无法理解如何为特定客户注册贷款。在尝试这样做时，PostMan 拒绝注册新贷款并显示以下消息：

{ “时间戳”：1504429329213， “状态”：400， "error": "错误请求", “异常”：“org.springframework.http.converter.HttpMessageNotReadableException”， “消息”：“JSON 解析错误：无法构造 com.reborn.xxx.backend.models.Client 的实例：没有 int/Int-argument 构造函数/工厂方法从数值 (1) 反序列化；嵌套异常是 com。 fastxml.jackson.databind.JsonMappingException：无法构造 com.reborn.xxx.backend.models.Client 的实例：没有 int/Int-argument 构造函数/工厂方法从数字值 (1)\n 在 [来源：java .io.PushbackInputStream@121b34b; line: 3, column: 12] (通过引用链: com.reborn.xxx.backend.models.Loan[\"client\"])", “路径”：“/api/loans/add” }

LoanController：

@RestController
@RequestMapping(value = "/api/loans")
public class LoanController extends BaseController {

    @Autowired
    private LoanService loanService;

    @RequestMapping(value = "/add", method = RequestMethod.POST, consumes = MediaType.APPLICATION_JSON_VALUE)
    public ResponseEntity registerLoan(@RequestBody Loan loan) {
        Loan regLoan = loanService.registerLoan(loan);
        if(regLoan == null) {
            return new ResponseEntity(HttpStatus.INTERNAL_SERVER_ERROR);
        }
        return new ResponseEntity(HttpStatus.CREATED);
    }
}

有什么想法可以实现这个目标吗？

更新1：

客户端存储库：

@Repository
public interface ClientRepository extends JpaRepository<Client, Long>{
}

LoanRepository：

@Repository
public interface LoanRepository extends JpaRepository<Loan, Long> {
}

JSON 添加客户端（有效）：

{
"id": 1,
"name": "Arthur",
"surname": "Doyle",
"phone": 777458642,
"email": "adoyle@imperial.com"
}

JSON 向特定客户提供贷款（失败）：

{
    "id": 1,
    "client": "http://loacalhost:8080/api/clients/find/1",
    "startDate": 20170902,
    "originTerm": 20170902,
    "maxPossibleAmount": 5555.0000,
    "notes": null
}

Answer 1

您的Loan 与Client 相关联。因此，如果您想为客户创建贷款，请使用此有效负载：

POST http://loacalhost:8080/api/loans 
{
    "originTerm": ...
    "maxPossibleAmount": ...
    "notes": ...
    "client": "http://loacalhost:8080/api/clients/1"
}

不需要自定义控制器。

附：将@JsonIgnoreProperties("loans") 添加到Loan.client 以避免stackoverflow 异常...

附言在@OneToMany 中，fetch 参数默认为FetchType.LAZY，因此您可以避免它。