如何使用 JPA 持久化 LocalDate？

Question

我想將沒有時間的日期存儲到我的數據庫中。 所以，我選擇使用LocalDate類型。

如本文所述，我使用 JPA 轉換器將LocalDate轉換為Date 。

但是，當我想保留我的實體（使用 POST 和 PUT 請求）時，我遇到了一些麻煩。

錯誤

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]

代碼

轉換器

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());
    }
}

實體

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));

感謝幫助。

編輯 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"]>

Answer 1

使用 JPA 2.2，您不再需要使用轉換器，它添加了對以下 java.time 類型映射的支持：

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;

Answer 2

JPA 2.2 支持LocalDate ，因此不需要轉換器。

Hibernate 從 5.3 版本開始也支持它。

查看這篇文章了解更多詳情。

Answer 3

Hibernate 5 支持 java 8，因此您可以將其添加到 pom.xml：

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

這為您提供了開箱即用的LocalDate和LocalDateTime映射。

Answer 4

JPA 2.2增加了對映射Java 8日期/時間API，像支持LocalDate ， LocalTime ， LocalDateTime ， OffsetDateTime或OffsetTime 。

因此，讓我們假設我們有以下實體：

@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
}

請注意， subscribedOn屬性是一個LocalDate Java 對象。

持久化UserAccount ：

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

entityManager.persist(user);

Hibernate 生成正確的 SQL INSERT 語句：

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

在獲取UserAccount實體時，我們可以看到從數據庫中正確獲取了LocalDate ：

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

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

Answer 5

我認為您可以編寫自己的轉換器，請檢查答案： Spring Data JPA - 從字符串轉換日期和/或時間時轉換失敗