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.Date
或java.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);
}
}
和
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/