Question

在休眠状态下，有什么方法可以生成长类型的自定义ID作为主键吗？我已经阅读过此hibernate ID，但是当Id的类型为String时，它可以工作。我想要的ID看起来像“ yyyymm00001”，例如：20180900001;在这里，下面的方法返回我想要的格式，但是是String类型。如何将其设置为长型？

@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

要更正此问题，必须在实体中添加此注释，并为@Id设置Long或Integer类型。

myData.write.format('jdbc').options(
      url='jdbc:%s' % url,
      driver='org.postgresql.Driver',
      dbtable='pyspark_user',
      user='postgres',
      password='').mode('append').save()

知道这一点

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

在休眠生成器类中，将此转换添加到Long

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

而不是：

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

让我知道一切是否好

休眠中的自定义Long类型ID生成

1 个答案: