我需要访问一个现有的MySQL数据库,它使用BIGINT列来存储时间戳:

create table mytable (created bigint);

现在我更喜欢使用java.util.Datejava.time.Instant实例而不是整数,所以我尝试让hibernate直接转换值。不幸的是,hibernate在进行如下注释时无法识别列:
@Column(name = "created")
private Date created;

或者像这样:
@Column(name = "created")
@Temporal(TemporalType.TIMESTAMP)
private Calendar created;

这确实会在未来返回一些东西,比如2017-07-01T04:14:00+02:00。如何使hibernate正确地转换BIGINT列,从而不必在getter和setter中转换它们?

最佳答案

hibernate的TemporalType.TIMESTAMP映射到java.sql.Timestamp。构造函数采用长值,即自纪元以来的毫秒数。数据库中的值存储为unix时间戳,它是自纪元以来的秒数。
我写了一个需要几秒钟而不是几毫秒的Timestamp并创建了UnixTimestampType实例:

public class UnixTimestampType extends AbstractSingleColumnStandardBasicType<Date> implements IdentifierType<Date>, LiteralType<Date> {
    private static final long serialVersionUID = 1L;
    public static final UnixTimestampType INSTANCE = new UnixTimestampType();

    public UnixTimestampType() {
        super(UnixTimestampTypeDescriptor.INSTANCE, JdbcDateTypeDescriptor.INSTANCE);
    }

    @Override
    public String getName() {
        return "date";
    }

    @Override
    public String[] getRegistrationKeys() {
        return new String[] { getName(), java.sql.Date.class.getName() };
    }

    @Override
    public String objectToSQLString(Date value, Dialect dialect) throws Exception {
        final java.sql.Date jdbcDate = java.sql.Date.class.isInstance(value) ? (java.sql.Date) value : new java.sql.Date(value.getTime());
        return StringType.INSTANCE.objectToSQLString(jdbcDate.toString(), dialect);
    }

    @Override
    public Date stringToObject(String xml) {
        return fromString(xml);
    }
}


public class UnixTimestampTypeDescriptor implements SqlTypeDescriptor {
    private static final long serialVersionUID = 1L;
    public static final UnixTimestampTypeDescriptor INSTANCE = new UnixTimestampTypeDescriptor();

    @Override
    public int getSqlType() {
        return Types.INTEGER;
    }

    @Override
    public boolean canBeRemapped() {
        return true;
    }

    @Override
    public <X> ValueBinder<X> getBinder(final JavaTypeDescriptor<X> javaTypeDescriptor) {
        return new BasicBinder<X>(javaTypeDescriptor, this) {
            @Override
            protected void doBind(PreparedStatement st, X value, int index, WrapperOptions options) throws SQLException {
                Date date = javaTypeDescriptor.unwrap(value, Date.class, options);
                date.setTime(date.getTime() / 1000);
                st.setDate(index, date);
            }
        };
    }

    @Override
    public <X> ValueExtractor<X> getExtractor(final JavaTypeDescriptor<X> javaTypeDescriptor) {
        return new BasicExtractor<X>(javaTypeDescriptor, this) {
            @Override
            protected X doExtract(ResultSet rs, String name, WrapperOptions options) throws SQLException {
                Date date = new Date(rs.getLong(name) * 1000);
                return javaTypeDescriptor.wrap(date, options);
            }
        };
    }
}

关于java - 在JPA/Hibernate中将BIGINT列用于java.util.Date或java.time.Instant,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/13647030/

10-08 22:34
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