Spring 启动：错误：在响应提交后无法调用 sendError()

Question

我收到这个错误。 提交响应后无法调用 sendError()有人可以帮我找出原因吗？

    @Entity
    public class Product {

        @Id
        @GeneratedValue(strategy = GenerationType.IDENTITY)
        private int id;

        @OneToOne(
                fetch = FetchType.LAZY,
                cascade = CascadeType.ALL
        )
        @JoinColumn(name = "details_id")
        private Details details;
//Getters and setters left out for brevity
    }


@Entity
public class Details {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int id;
    private String name;
    private String description;
    private float price;
    private float discount;
    @OneToOne(mappedBy = "details")
    private Product product;
}


@RestController
public class ProductController {
    @Autowired
    ProductRepository productRepository;

    @GetMapping("/getAllProducts")
    public Iterable<Product> getAllProducts(){
        return productRepository.findAll();
    }
}

   @RestController
public class DetialsController {
    @Autowired
    ProductRepository productRepository;

    @Autowired
    DetailsRepository detailsRepository;

    @PostMapping("/details")
    public Details addDetails(@RequestBody Details details) {
        Product newProduct = new Product();
        newProduct.setDetails(details);
        productRepository.save(newProduct);
        return detailsRepository.save(details);
    }
}

我可以对 /details 进行 POST 调用； 成功添加详细信息。 但是当我对 /getAllProducts 进行 GET 调用时，我收到此错误无法在响应提交后调用 sendError()

Answer 1

这是双向关系的问题，因为它们相互引用，反序列化时，Jackson无限循环运行。 我的第一个建议是在关系的一端添加@JsonIgnore 。

@OneToOne(mappedBy = "details")
@JsonIgnore
private Product product;

之后，如果这解决了您的问题，则可以查看@ JsonManagedReference / @ JsonBackReference和@JsonIdentityInfo。

您也可以查看此链接以获得更多信息

Answer 2

你可以使用这个：

    @Entity
    public class Product {

        @Id
        @GeneratedValue(strategy = GenerationType.IDENTITY)
        private int id;


        @JsonBackReference(value = "details_id")
        @OneToOne(
                fetch = FetchType.LAZY,
                cascade = CascadeType.ALL
        )
        @JoinColumn(name = "details_id")
        private Details details;
//Getters and setters left out for brevity
    }


@Entity
public class Details {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int id;
    private String name;
    private String description;
    private float price;
    private float discount;

    @JsonManagedReference(value = "details")
    @OneToOne(mappedBy = "details",,cascade=CascadeType.ALL)
    private Product product;
}