相关疑难解决方法(0)

如何将Jackson JSON映射器与Java 8 LocalDateTime一起使用？

org.codehaus.jackson.map.JsonMappingException:无法从JSON String实例化类型[simple type,class java.time.LocalDateTime]的值; 没有单字符串构造函数/工厂方法(通过引用链:MyDTO ["field1"] - > SubDTO ["date"])

我的json序列化有问题ZonedDateTime.当转换为json时,它产生了一个巨大的对象,我不希望每次都传输所有这些数据.所以我尝试将其格式化为ISO,但它不起作用.我怎样才能让它格式化？

这是我的实体类:

@MappedSuperclass
public abstract class AuditBase {

    @Id
    @GeneratedValue
    private Long id;

    @CreatedDate
    private ZonedDateTime createdDate;

    @LastModifiedDate
    private ZonedDateTime lastModifiedDate;

    @DateTimeFormat(iso = DateTimeFormat.ISO.DATE_TIME)
    public ZonedDateTime getLastModifiedDate() {
        return lastModifiedDate;
    }

    public void setLastModifiedDate(ZonedDateTime lastModifiedDate) {
        this.lastModifiedDate = lastModifiedDate;
    }

    @DateTimeFormat(iso = DateTimeFormat.ISO.DATE_TIME)
    public ZonedDateTime getCreatedDate() {
        return createdDate;
    }

    public void setCreatedDate(ZonedDateTime createdDate) {
        this.createdDate = createdDate;
    }

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    @PrePersist
    public void …

我已经通过链接：JSON 解析错误：无法构造 java.time.LocalDate 的实例：没有从字符串值反序列化的字符串参数构造函数/工厂方法，但得到以下错误。

错误：

com.fasterxml.jackson.databind.exc.InvalidDefinitionException: Cannot construct instance of `java.time.Instant` (no Creators, like default construct, exist): no String-argument constructor/factory method to deserialize from String value ('2019-08-28T15:15:44.183Z')

爪哇

@Builder
@Data
@AllArgsConstructor
@NoArgsConstructor
@JsonIgnoreProperties(ignoreUnknown = true)
public class StudentResponse {
    private Integer a;
    private String b;
    private String c;
    private String d;
    private String e;
    private Instant lastUpdateDate;
}