时间戳

首页 > 解决方案 > 如何使用数据类在 Kotlin 中解决“无法编写 JSON：无限递归 (StackOverflowError)”？

json - 如何使用数据类在 Kotlin 中解决“无法编写 JSON：无限递归 (StackOverflowError)”？

问题描述

我知道该主题中有几篇文章，但我只是找不到使用 Kotlin 数据类的文章。所以我正在尝试使用 Spring Boot 在 Kotlin 中使用 H2 数据库制作一个 REST API，并且我也在使用 Postman。我的类的一些属性具有 List 类型。每次我尝试在 Postman 中为这些列表添加一些值然后尝试获取结果时，我都会收到以下错误：

在此处输入图像描述

我有三个班级：

食谱.kt：

@Entity
data class Recipe(
    @Id
    @SequenceGenerator(name = RECIPE_SEQUENCE, sequenceName = RECIPE_SEQUENCE, initialValue = 1, allocationSize = 1)
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    var id: Long = 0,
    val name: String,
    var cookTime: String?,
    var servings: String?,
    var directions: String?,
    @OneToMany(cascade = [CascadeType.ALL], mappedBy = "recipe")
    @JsonManagedReference
    var ingredient: List<Ingredient>?,
    @ManyToMany
    @JsonManagedReference
    @JoinTable(
        name = "recipe_category",
        joinColumns = [JoinColumn(name = "recipe_id")],
        inverseJoinColumns = [JoinColumn(name = "category_id")]
    )
    var category: List<Category>?,
    @Enumerated(value = EnumType.STRING)
    var difficulty: Difficulty?
    ) { companion object { const val RECIPE_SEQUENCE: String = "RECIPE_SEQUENCE" } }

类别.kt

    @Entity
data class Category(
    @Id
    @SequenceGenerator(name = CATEGORY_SEQUENCE, sequenceName = CATEGORY_SEQUENCE, initialValue = 1, allocationSize = 1)
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    val id: Long = 0,
    var name: String,
    @ManyToMany(mappedBy = "category")
    @JsonBackReference
    var recipe: List<Recipe>?
) { companion object { const val CATEGORY_SEQUENCE: String = "CATEGORY_SEQUENCE" } }

成分.kt

    @Entity
data class Ingredient(
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    val id: Long = 0,
    var description: String?,
    var amount: BigDecimal?,
    @ManyToOne
    @JsonBackReference
    var recipe: Recipe?,
    var unitOfMeasure: String?
)

食谱响应.kt

data class RecipeResponse (var id:Long,
                           var name:String,
                           var cookTime:String?,
                           var servings:String?,
                           var directions:String?,
                           var ingredient:List<Ingredient>?,
                           var category: List<Category>?,
                           var difficulty: Difficulty?)

食谱资源.kt

@RestController
@RequestMapping(value = [BASE_RECIPE_URL])
class RecipeResource(private val recipeManagementService: RecipeManagementService)
{

    @GetMapping
    fun findAll(): ResponseEntity<List<RecipeResponse>> = ResponseEntity.ok(this.recipeManagementService.findAll())

食谱管理服务.kt

@Service
class RecipeManagementService (@Autowired private val recipeRepository: RecipeRepository,
                               private val addRecipeRequestTransformer: AddRecipeRequestTransformer) {

    fun findAll(): List<RecipeResponse> = this.recipeRepository.findAll().map(Recipe::toRecipeResponse)

标签： jsonspring-bootkotlinpostmanh2

解决方案

您必须在@ManyToOne 和 @OneToMany 字段之上添加@JsonIgnore 。Spring 将忽略返回该对象的那些字段。外汇。

@Entity
data class Ingredient(
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    val id: Long = 0,
    var description: String?,
    var amount: BigDecimal?,
    @ManyToOne
    @JsonIgnore
    var recipe: Recipe?, // This field will not be returned in JSON response.
    var unitOfMeasure: String?
)

在这里，还要注意是否要在响应中包含此 ManyToOne 或 OneToMany 关系的某些字段。您必须使用 ObjectNode 制定您的响应，然后将其返回。

编辑 ：


ObjectMapper objectMapper = new ObjectMapper();

@RestController
public ResponseEntity<String> someFunction() throws Exception {
    ObjectNode msg = objectMapper.createObjectNode();
    msg.put("success", true);
    msg.put("field1", object.getValue1());
    msg.put("field2", object.getValue2());
    return ResponseEntity.ok()
        .header("Content-Type", "application/json")
        .body(objectMapper.writerWithDefaultPrettyPrinter()
        .writeValueAsString(res));
}

这里的代码是java，你可以把它转换成Kotlin，我认为它会是一样的。在这里，您可以编写自己的对象名称 ex，而不是 object。成分.getDescription() 并且可以生成响应。

推荐阅读