How to use the Java 8 LocalDateTime with JPA and Hibernate

32,839

Solution 1

Since Hibernate 4 doesn't support it you need to implement a user type as shown in this example.

import org.hibernate.HibernateException;
import org.hibernate.engine.spi.SessionImplementor;
import org.hibernate.type.StandardBasicTypes;
import org.hibernate.usertype.EnhancedUserType;

import java.io.Serializable;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.sql.Types;
import java.time.Instant;
import java.time.LocalDateTime;
import java.time.ZoneId;
import java.util.Date;

public class LocalDateTimeUserType implements EnhancedUserType, Serializable {

    private static final int[] SQL_TYPES = new int[]{Types.TIMESTAMP};

    @Override
    public int[] sqlTypes() {
        return SQL_TYPES;
    }

    @Override
    public Class returnedClass() {
        return LocalDateTime.class;
    }

    @Override
    public boolean equals(Object x, Object y) throws HibernateException {
        if (x == y) {
            return true;
        }
        if (x == null || y == null) {
            return false;
        }
        LocalDateTime dtx = (LocalDateTime) x;
        LocalDateTime dty = (LocalDateTime) y;
        return dtx.equals(dty);
    }

    @Override
    public int hashCode(Object object) throws HibernateException {
        return object.hashCode();
    }


    @Override
    public Object nullSafeGet(ResultSet resultSet, String[] names, SessionImplementor session, Object owner)
            throws HibernateException, SQLException {
        Object timestamp = StandardBasicTypes.TIMESTAMP.nullSafeGet(resultSet, names, session, owner);
        if (timestamp == null) {
            return null;
        }
        Date ts = (Date) timestamp;
        Instant instant = Instant.ofEpochMilli(ts.getTime());
        return LocalDateTime.ofInstant(instant, ZoneId.systemDefault());
    }

    @Override
    public void nullSafeSet(PreparedStatement preparedStatement, Object value, int index, SessionImplementor session)
            throws HibernateException, SQLException {
        if (value == null) {
            StandardBasicTypes.TIMESTAMP.nullSafeSet(preparedStatement, null, index, session);
        } else {
            LocalDateTime ldt = ((LocalDateTime) value);
            Instant instant = ldt.atZone(ZoneId.systemDefault()).toInstant();
            Date timestamp = Date.from(instant);
            StandardBasicTypes.TIMESTAMP.nullSafeSet(preparedStatement, timestamp, index, session);
        }
    }

    @Override
    public Object deepCopy(Object value) throws HibernateException {
        return value;
    }

    @Override
    public boolean isMutable() {
        return false;
    }

    @Override
    public Serializable disassemble(Object value) throws HibernateException {
        return (Serializable) value;
    }

    @Override
    public Object assemble(Serializable cached, Object value) throws HibernateException {
        return cached;
    }

    @Override
    public Object replace(Object original, Object target, Object owner) throws HibernateException {
        return original;
    }

    @Override
    public String objectToSQLString(Object object) {
        throw new UnsupportedOperationException();
    }

    @Override
    public String toXMLString(Object object) {
        return object.toString();
    }

    @Override
    public Object fromXMLString(String string) {
        return LocalDateTime.parse(string);
    }

}

The new usertype can then be used in the mapping with the @Type annotation. For e.g.

@Type(type="com.hibernate.samples.type.LocalDateTimeUserType")
@Column(name = "invalidate_token_date")
private LocalDateTime invalidateTokenDate;

The @Type annotation needs a full path to the class that implements the userType interface; this is the factory for producing the target type of the mapped column.

Here's how to do the same thing in JPA2.1

Solution 2

For any Hibernate 5.x users, there is

<dependency>
    <groupId>org.hibernate</groupId>
    <artifactId>hibernate-java8</artifactId>
    <version>5.0.0.Final</version>
</dependency>

You don't need to do anything else. Just add the dependency, and the Java 8 time types should work like any other basic types, no annotations required.

private LocalDateTime invalidateTokenDate;

Note: this won't save to timestamp type though. Testing with MySQL, it saves to datetime type.

Solution 3

Since version 2.2, JPA offers support for mapping Java 8 Date/Time API, like LocalDateTime, LocalTime, LocalDateTimeTime, OffsetDateTime or OffsetTime.

Also, even with JPA 2.1, Hibernate 5.2 supports all Java 8 Date/Time API by default.

In Hibernate 5.1 and 5.0, you have to add the hibernate-java8 Maven dependency.

So, let's assume we have the following entity:

@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 LocalDateTime subscribedOn;

    //Getters and setters omitted for brevity
}

Notice that the subscribedOn attribute is a LocalDateTime Java object.

When persisting the UserAccount:

UserAccount user = new UserAccount()
    .setId(1L)
    .setFirstName("Vlad")
    .setLastName("Mihalcea")
    .setSubscribedOn(
        LocalDateTime.of(
            2020, 5, 1,
            12, 30, 0
        )
    );

entityManager.persist(user);

Hibernate generates the proper SQL INSERT statement:

INSERT INTO user_account (
    first_name, 
    last_name, 
    subscribed_on, 
    id
) 
VALUES (
    'Vlad', 
    'Mihalcea', 
    '2020-05-01 12:30:00.0', 
    1
)

When fetching the UserAccount entity, we can see that the LocalDateTime is properly fetched from the database:

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

assertEquals(
    LocalDateTime.of(
        2020, 5, 1,
        12, 30, 0
    ),
    userAccount.getSubscribedOn()
);

Solution 4

If you can use Java EE 7, there is more elegant solution:

>> Implement this:

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

    @Override
    public Date convertToDatabaseColumn(LocalDateTime date) {
        if (date == null){
            return null;
        }
        return date.toDate();
    }

    @Override
    public LocalDateTime convertToEntityAttribute(Date value) {
        if (value == null) {
            return null;
        }
        return LocalDateTime.fromDateFields(value);
    }
} 

>> Use like this:

... 
@Column(name = "invalidate_token_date")
private LocalDateTime invalidateTokenDate;
....

Value (autoApply = true) means that @Converter is automatically used for conversion of every LocalDateTime property in your JPA Entity.

Btw, AttributeConverter is pretty good for mapping Enums too.

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Updated on January 09, 2021

Comments

  • gstackoverflow
    gstackoverflow over 3 years

    I have the following class description snippet:

    ... 
    @Column(name = "invalidate_token_date")
    @Temporal(TemporalType.TIMESTAMP)
    private LocalDateTime invalidateTokenDate;
    ....
    

    This code doesn't work on Hibernate 4 because @Temporal doesn't support LocalDateTime.

    I saw the suggestion on how to use LocalDateTime from Joda-Time but I use Java 8.

  • gstackoverflow
    gstackoverflow over 9 years
    And how should I change mapping after adding class from you answer?
  • obesechicken13
    obesechicken13 almost 9 years
    Which library gives you the localDateTime toDate and fromDateFields methods?
  • obesechicken13
    obesechicken13 almost 9 years
  • Quentin Klein
    Quentin Klein over 8 years
    Saved my night :) Thanks
  • Radosław Osiński
    Radosław Osiński over 8 years
    That is not enough for spring configuration. After adding hibernate-java8 dependency to project ClassCastException still occurs.
  • Peter DeGregorio
    Peter DeGregorio about 8 years
    Thank you. Adding hibernate-java8 artifact worked for me with Spring 4.2.5 and Hibernate 5.1.31. Was getting org.hibernate.type.SerializationException: could not deserialize on the entity with LocalDateTime fields.
  • Jeroen van Dijk-Jun
    Jeroen van Dijk-Jun about 7 years
    The question is about Hibernate4, so this is a good idea, but no answer for this question
  • Bruno Gasparotto
    Bruno Gasparotto over 6 years
    It's actually a JPA feature, which also works with Spring Data JPA, not only with Java EE.
  • Gwaptiva
    Gwaptiva over 6 years
    Note that if you convert this to store java.util.LocalDate fields, you cannot go via the java.sql.Date.toInstant() method, as that throws an exception; use toLocalDate instead.
  • Viktor Vix Jančík
    Viktor Vix Jančík about 6 years
    No longer needed in Hibernate 5.3
  • deFreitas
    deFreitas over 5 years
    You sir saved my day, working perfectly on postgres 10 timestamp type