java - hibernate中自定义Long类型ID生成

Question

有没有办法在hibernate中生成长类型的自定义id作为主键？我读过这篇hibernate ID但当 Id 类型为 String 时，它可以工作。我想要的 id 看起来像“yyyymm00001”，例如:20180900001；下面的方法返回我想要的格式，但是是字符串类型。我怎样才能把它变成长类型？

@Override
    public Serializable generate(SharedSessionContractImplementor session, Object object) throws HibernateException {
           Serializable result = null;
            Connection connection = null;
            Statement statement = null;
            ResultSet resultSet = null;


            String year= String.valueOf(Calendar.getInstance().get(Calendar.YEAR));
            String month= String.format("%02d",Calendar.getInstance().get(Calendar.MONTH)+1);
            try {
                connection = session.connection();
                statement = connection.createStatement();
                try {

                     statement.executeUpdate("UPDATE " + DEFAULT_SEQUENCE_NAME + " SET next_val=LAST_INSERT_ID(next_val+1)");
                    resultSet = statement.executeQuery("SELECT next_val FROM  " + DEFAULT_SEQUENCE_NAME);
                } catch (Exception e) {

                    System.out.println("In catch, cause : Table is not available.");

                    statement.executeUpdate("UPDATE " + DEFAULT_SEQUENCE_NAME + " SET next_val=LAST_INSERT_ID(next_val+1)");
                    resultSet = statement.executeQuery("SELECT next_val FROM  " + DEFAULT_SEQUENCE_NAME);

                }
                if (resultSet.next()) {

                    int nextValue = resultSet.getInt(1);
                    String suffix = String.format("%05d", nextValue);
                    result = year.concat(month).concat(suffix);
                    System.out.println("Custom generated sequence is : " + result);
                }
            } catch (SQLException e) {
                e.printStackTrace();
            }
            return result;
    }

Answer 1

要解决此问题，您必须在实体中添加此注释并为 @Id 设置 Long 或 Integer 类型

@Id
@Column(name = Contact.ENCOIDFSYS)
@GeneratedValue(generator = IdfsysGeneratorDefinition.NAME)
@GenericGenerator(name = IdfsysGeneratorDefinition.NAME, strategy = IdfsysGeneratorDefinition.CLASS_NAME, parameters = {  })
private Integer id;

知道

public static final String CLASS_NAME = "com.sybaway.generators.hibernate.HibernateIdfsysGenerator";
public static final String NAME = "idfsysGenerator";

并在 hibernate 生成器类中，将此转换添加到 Long

result = Long.parseLong(year.concat(month).concat(suffix));

而不是:

result = year.concat(month).concat(suffix);

如果一切顺利请告诉我