如何使用 JPA 保留 LocalDate？

Question

我想将没有时间的日期存储到我的数据库中。所以，我选择使用 LocalDate 类型。

如 本文所述，我使用 JPA 转换器将 LocalDate 转换为 Date 。

但是，当我想要保留我的实体（使用 POST 和 PUT 请求）时，我遇到了一些麻烦。

错误

2019-02-23 11:26:30.254  WARN 2720 --- [-auto-1-exec-10] .w.s.m.s.DefaultHandlerExceptionResolver : Resolved [org.springframework.http.converter.HttpMessageNotReadableException: JSON parse error: Expected array or string.; nested exception is com.fasterxml.jackson.databind.exc.MismatchedInputException: Expected array or string.
 at [Source: (PushbackInputStream); line: 1, column: 104] (through reference chain: ...entity.MyObject["startdate"])]

org.springframework.http.converter.HttpMessageConversionException: Type definition error: [simple type, class org.springframework.http.ResponseEntity]; nested exception is com.fasterxml.jackson.databind.exc.InvalidDefinitionException: Cannot construct instance of `org.springframework.http.ResponseEntity` (no Creators, like default construct, exist): cannot deserialize from Object value (no delegate- or property-based Creator)
 at [Source: (PushbackInputStream); line: 1, column: 2]

代码

转换器

package ...entity;

import javax.persistence.AttributeConverter;
import javax.persistence.Converter;
import java.time.LocalDate;
import java.sql.Date;

@Converter(autoApply = true)
public class LocalDateAttributeConverter implements AttributeConverter<LocalDate, Date> {

    @Override
    public Date convertToDatabaseColumn(LocalDate locDate) {
        return (locDate == null ? null : Date.valueOf(locDate));
    }

    @Override
    public LocalDate convertToEntityAttribute(Date sqlDate) {
        return (sqlDate == null ? null : sqlDate.toLocalDate());
    }
}

实体

package ...entity;

import org.hibernate.annotations.ColumnDefault;

import javax.persistence.*;
import java.time.LocalDate;
import java.util.HashSet;
import java.util.Set;

@Entity
public class MyObject {

    @Id
    private String id;
    private LocalDate startdate;
    private LocalDate enddate;

    public MyObject() {}

    public MyObject(LocalDate enddate) {
        this.startdate = LocalDate.now();
        this.enddate = enddate;
    }

    ...
}

“主要的”

 private DateTimeFormatter formatter = DateTimeFormatter.ofPattern("yyyy-MM-dd");
MyObject myobject = new MyObject(LocalDate.parse("2019-03-01", formatter));

感谢帮助。

编辑 1：打印 MyObject

  HttpHeaders headers = new HttpHeaders();
 headers.setContentType(MediaType.APPLICATION_JSON);
 HttpEntity<String> entity = new HttpEntity<>(this.toJsonString(myObject), headers);
 System.out.println(entity.toString());

 // <{"id":"ba6649e4-6e65-4f54-8f1a-f8fc7143b05a","startdate":{"year":2019,"month":"FEBRUARY","dayOfMonth":23,"dayOfWeek":"SATURDAY","era":"CE","dayOfYear":54,"leapYear":false,"monthValue":2,"chronology":{"id":"ISO","calendarType":"iso8601"}},"enddate":{"year":2019,"month":"MARCH","dayOfMonth":1,"dayOfWeek":"FRIDAY","era":"CE","dayOfYear":60,"leapYear":false,"monthValue":3,"chronology":{"id":"ISO","calendarType":"iso8601"}}},[Content-Type:"application/json"]>

原文由 Royce 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

JPA 2.2 支持 LocalDate ，因此不需要转换器。

从 5.3 版本开始，Hibernate 也支持它。

查看 这篇文章 了解更多详情。

原文由 Peter Šály 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 2

使用 JPA 2.2，您不再需要使用转换器，它添加了对以下 java.time 类型映射的支持：

 java.time.LocalDate
java.time.LocalTime
java.time.LocalDateTime
java.time.OffsetTime
java.time.OffsetDateTime

 @Column(columnDefinition = "DATE")
private LocalDate date;
@Column(columnDefinition = "TIMESTAMP")
private LocalDateTime dateTime;
@Column(columnDefinition = "TIME")
private LocalTime localTime;

原文由 Michael Mesfin 发布，翻译遵循 CC BY-SA 4.0 许可协议