Question

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

如此处https://thoughts-on-java.org/persist-localdate-localdatetime-jpa/所述，我使用转换器将LocalDate转换为Date。

但是，当我要保留我的实体（带有POST和PUT请求）时，会遇到一些麻烦。

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”

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：MyObject的打印

 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"]>

Answer 1

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

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;

Answer 2

由于这是一个非常常见的问题，所以此答案基于this article，我写了用JPA映射日期和时间戳的最佳方法。

JPA 2.2添加了对映射Java 8 Date/Time API的支持，例如LocalDate，LocalTime，LocalDateTime，OffsetDateTime或OffsetTime。

因此，假设我们具有以下实体：

@Entity(name = "UserAccount")
@Table(name = "user_account")
public class UserAccount {

    @Id
    private Long id;

    @Column(name = "first_name", length = 50)
    private String firstName;

    @Column(name = "last_name", length = 50)
    private String lastName;

    @Column(name = "subscribed_on")
    private LocalDate subscribedOn;

    //Getters and setters omitted for brevity
}

请注意，subscribedOn属性是一个LocalDate Java对象。

坚持UserAccount时：

UserAccount user = new UserAccount()
    .setId(1L)
    .setFirstName("Vlad")
    .setLastName("Mihalcea")
    .setSubscribedOn(
        LocalDate.of(
            2013, 9, 29
        )
    );

entityManager.persist(user);

Hibernate生成正确的SQL INSERT语句：

INSERT INTO user_account (
    first_name, 
    last_name, 
    subscribed_on, 
    id
) 
VALUES (
    'Vlad', 
    'Mihalcea', 
    '2013-09-29', 
    1
)

在获取UserAccount实体时，我们可以看到从数据库中正确获取了LocalDate：

UserAccount userAccount = entityManager.find(
    UserAccount.class, 1L
);

assertEquals(
    LocalDate.of(
        2013, 9, 29
    ),
    userAccount.getSubscribedOn()
);

Answer 3

使用JPA 2.2不需要转换器。 Hibernate从5.3版本开始支持它。 https://vladmihalcea.com/whats-new-in-jpa-2-2-java-8-date-and-time-types/

Answer 4

Hibernate 5支持Java 8，因此您可以将其添加到pom.xml：

<dependency>
    <groupId>org.hibernate</groupId>
    <artifactId>hibernate-java8</artifactId>
    <version>5.1.0.Final</version>
</dependency>

这为您提供了LocalDate和LocalDateTime的现成映射。

Answer 5

我认为您可以编写自己的Converter，请检查答案：Spring Data JPA - Conversion failed when converting date and/or time from character string

如何使用JPA保留LocalDate？

5 个答案: