java - 如何通过读取 JSON 来修复 UnrecognizedPropertyException？

Question

我有一个实体类:

@Entity
@Table(name = "meals")
public class Meal extends AbstractNamedEntity implements Serializable {
    private static final long serialVersionUID = 1L;

    @NotNull(message = "Price must be added")
    private Integer price;

    @JsonIgnore
    @JsonProperty(access = JsonProperty.Access.READ_ONLY)
    @DateTimeFormat(pattern = "yyyy-MM-dd")
    private LocalDate date = LocalDate.now();

    @JsonIgnore
    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "restaurant_id", nullable = false)
    @OnDelete(action = OnDeleteAction.CASCADE)
    private Restaurant restaurant;

    public Meal() {
    }
    //getters and setters
}

当我使用测试数据测试我的 Controller 时:

public class MealTestData {
    private static final Integer START_SEQ = 100000;

    public static final Meal MEAL_1 = new Meal(START_SEQ + 6, "ChickenBurger set", 300, RESTAURANT_1);
    public static final Meal MEAL_2 = new Meal(START_SEQ + 7, "CheeseBurger set", 400, RESTAURANT_1);
    public static final Meal MEAL_3 = new Meal(START_SEQ + 8, "FishBurger set", 500, RESTAURANT_1);
}

我得到这个异常:

java.lang.IllegalArgumentException: Invalid read array from JSON: '[{"id":100006,"name":"ChickenBurger set","price":300,"date":"2020-01-26"},{"id":100007,"name":"CheeseBurger set","price":400,"date":"2020-01-26"},{"id":100008,"name":"FishBurger set","price":500,"date":"2020-01-26"}]';

Caused by: com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "date" (class ru.topjava.graduation.model.Meal), not marked as ignorable (3 known properties: "price", "id", "name"]) at [Source: (String)"[{"id":100006,"name":"ChickenBurger set","price":300,"date":"2020-01-26"},{"id":100007,"name":"CheeseBurger set","price":400,"date":"2020-01-26"},{"id":100008,"name":"FishBurger set","price":500,"date":"2020-01-26"}]"; line: 1, column: 62] (through reference chain: ru.topjava.graduation.model.Meal["date"])

是什么原因以及如何解决？我不能在测试中忽略该字段

提供了“日期”字段的 getter 和 setter:

public LocalDate getDate() {
    return date;
}

public void setDate(LocalDate date) {
    this.date = date;
}

Answer 1

您在“日期”字段上有 @JsonIgnore 注释:

@JsonIgnore
@JsonProperty(access = JsonProperty.Access.READ_ONLY)
@DateTimeFormat(pattern = "yyyy-MM-dd")
private LocalDate date = LocalDate.now();

但该字段存在于输入 json 中:

{"id":100006,"name":"ChickenBurger set","price":300,"date":"2020-01-26"},

因此，要么删除默认值生成，要么切换 ObjectMapper 的设置，如下所示:

mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);