Выборка данных с использованием спящего режима с Netbeans

Question

Я пытаюсь использовать Hibernate для получения данных из базы данных MySQL. Для достижения этой цели я создал:

1. Файл конфигурации Hibernate:

   <?xml version='1.0' encoding='utf-8'?>
   <!DOCTYPE hibernate-configuration PUBLIC
   "-//Hibernate/Hibernate Configuration DTD//EN" "http://www.hibernate.org/dtd/hibernate-configuration-3.0.dtd">
   <hibernate-configuration>
       <session-factory>
           <property name="hibernate.connection.driver_class">
               com.mysql.jdbc.Driver
           </property>
           <property name="hibernate.connection.url">
               jdbc:mysql://localhost:3306/employeesDbAF
           </property>
   
           <property name="hibernate.connection.username">root</property>
           <property name="hibernate.connection.password"></property>
           <property name="hibernate.connection.pool_size">50</property>
           <property name="show_sql">true</property>
           <property name="dialect">
               org.hibernate.dialect.MySQLDialect
           </property>
           <property name="hibernate.hbm2ddl.auto">validate</property>
   
   
       <mapping class="employeesapp.Employee"></mapping>    
       </session-factory>
   </hibernate-configuration>
   

2. Класс полезности

   public class HibernateUtil {
   
       private static SessionFactory sessionFactory;
       private static ServiceRegistry serviceRegistry;
   
       public static SessionFactory createSessionFactory() {
           Configuration configuration = new Configuration();
           configuration.configure();
           serviceRegistry = new StandardServiceRegistryBuilder().applySettings(
                   configuration.getProperties()).build();
           //sessionFactory = configuration.buildSessionFactory(serviceRegistry);
           sessionFactory = configuration.buildSessionFactory();
           return sessionFactory;
       }
   
       public static void close() {
           StandardServiceRegistryBuilder.destroy(serviceRegistry);
   
       }
   
   }
   

3. Основной класс:

открытый класс EmployeesApp расширяет javax.swing.JFrame {

public static void main(String[] args) {        

    Session session = HibernateUtil.createSessionFactory().openSession();
    Transaction tx = null;

    try {
        tx = session.beginTransaction();
        Employee emp = (Employee)session.get(Employee.class, 2);
        System.out.println(emp);
        tx.commit();
    } catch (HibernateException e) {
        if (tx != null) {
            tx.rollback();
        }
        System.out.println(e);
    } finally {
        HibernateUtil.close();
    }
}

EDIT: 4. Класс работника:

    @Entity
    @Table(name="employee")    
    public class Employee { 
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name="employee_ID")
    private int employee_ID;
    @Column(name="name")
    private String name;
    @Column(name="age")
    private int age;
    @Column(name="address")
    private String address;
    @Column(name="salary")
    private int salary;

    public int getEmployee_ID() {return employee_ID;}

    public void setEmployee_ID(int employee_ID) {this.employee_ID = employee_ID; }    

    public String getName() {return name;    }

    public void setName(String name) {this.name = name;    }

    public int getAge() {return age;}

    public void setAge(int age) {this.age = age;    }

    public String getAddress() {return address;    }

    public void setAddress(String address) {this.address = address;    }

    public int getSalary() {return salary;    }

    public void setSalary(int salary) {this.salary = salary;    }

    public Employee(int employee_ID, String name, int age, String address, int salary) {
        //this.employee_ID = employee_ID;
        this.name = name;
        this.age = age;
        this.address = address;
        this.salary = salary;
    }

    public Employee(){
    }

}

Когда дело доходит до выполнения кода, я не получаю строки из базы данных, просто какое-то общее предложение SELECT (последняя строка). Что я делаю неправильно?

run:
Jun 05, 2019 10:53:26 PM org.hibernate.annotations.common.reflection.java.JavaReflectionManager <clinit>
INFO: HCANN000001: Hibernate Commons Annotations {4.0.4.Final}
Jun 05, 2019 10:53:26 PM org.hibernate.Version logVersion
INFO: HHH000412: Hibernate Core {4.3.1.Final}
Jun 05, 2019 10:53:26 PM org.hibernate.cfg.Environment <clinit>
INFO: HHH000206: hibernate.properties not found
Jun 05, 2019 10:53:26 PM org.hibernate.cfg.Environment buildBytecodeProvider
INFO: HHH000021: Bytecode provider name : javassist
Jun 05, 2019 10:53:26 PM org.hibernate.cfg.Configuration configure
INFO: HHH000043: Configuring from resource: /hibernate.cfg.xml
Jun 05, 2019 10:53:26 PM org.hibernate.cfg.Configuration getConfigurationInputStream
INFO: HHH000040: Configuration resource: /hibernate.cfg.xml
Jun 05, 2019 10:53:26 PM org.hibernate.cfg.Configuration doConfigure
INFO: HHH000041: Configured SessionFactory: null
Jun 05, 2019 10:53:26 PM org.hibernate.engine.jdbc.connections.internal.DriverManagerConnectionProviderImpl configure
WARN: HHH000402: Using Hibernate built-in connection pool (not for production use!)
Jun 05, 2019 10:53:26 PM org.hibernate.engine.jdbc.connections.internal.DriverManagerConnectionProviderImpl buildCreator
INFO: HHH000401: using driver [com.mysql.jdbc.Driver] at URL [jdbc:mysql://localhost:3306/employeesDbAF]
Jun 05, 2019 10:53:26 PM org.hibernate.engine.jdbc.connections.internal.DriverManagerConnectionProviderImpl buildCreator
INFO: HHH000046: Connection properties: {user=root, password=****}
Jun 05, 2019 10:53:26 PM org.hibernate.engine.jdbc.connections.internal.DriverManagerConnectionProviderImpl buildCreator
INFO: HHH000006: Autocommit mode: false
Jun 05, 2019 10:53:26 PM org.hibernate.engine.jdbc.connections.internal.DriverManagerConnectionProviderImpl configure
INFO: HHH000115: Hibernate connection pool size: 50 (min=1)
Jun 05, 2019 10:53:27 PM org.hibernate.dialect.Dialect <init>
INFO: HHH000400: Using dialect: org.hibernate.dialect.MySQLDialect
Jun 05, 2019 10:53:27 PM org.hibernate.engine.transaction.internal.TransactionFactoryInitiator initiateService
INFO: HHH000399: Using default transaction strategy (direct JDBC transactions)
Jun 05, 2019 10:53:27 PM org.hibernate.hql.internal.ast.ASTQueryTranslatorFactory <init>
INFO: HHH000397: Using ASTQueryTranslatorFactory
Jun 05, 2019 10:53:27 PM org.hibernate.tool.hbm2ddl.SchemaValidator validate
INFO: HHH000229: Running schema validator
Jun 05, 2019 10:53:27 PM org.hibernate.tool.hbm2ddl.SchemaValidator validate
INFO: HHH000102: Fetching database metadata
Jun 05, 2019 10:53:27 PM org.hibernate.tool.hbm2ddl.TableMetadata <init>
INFO: HHH000261: Table found: employeesDbAF.employee
Jun 05, 2019 10:53:27 PM org.hibernate.tool.hbm2ddl.TableMetadata <init>
INFO: HHH000037: Columns: [address, employee_id, name, salary, age]
**Hibernate: select employee0_.employee_ID as employee1_0_0_, employee0_.address as address2_0_0_, employee0_.age as age3_0_0_, employee0_.name as name4_0_0_, employee0_.salary as salary5_0_0_ from employee employee0_ where employee0_.employee_ID=?
employeesapp.Employee@53b7f657**

Answer 1

Вы должны переопределить метод public String toString() в классе Employee.

@Override
public String toString() {
    return "Employee{" +
            "employee_ID=" + employee_ID +
            ", name='" + name + '\'' +
            ", age=" + age +
            ", address='" + address + '\'' +
            ", salary=" + salary +
            '}';
}

System.out.println(emp); метод просто вывести на консоль возвращаемое значение метода toString ()

Employee@53b7f657

Внутренне этовызовет,

// java.io.PrintStream    
public void println(Object x) {
            String s = String.valueOf(x);
            synchronized (this) {
                print(s);
                newLine();
            }
        }

---

// java.lang.String
public static String valueOf(Object obj) {
    return (obj == null) ? "null" : obj.toString();
}

И реализация этого метода в классе объекта будет

// java.lang.Object    
public String toString() {
    return getClass().getName() + "@" + Integer.toHexString(hashCode());
}

Так что, если у вас нет 'custom' public String toString() в вашем классе это то, что вы увидите, потому что каждый класс неявно является подклассом Object .

Исключая Object, у которого нет суперкласса, у каждого класса есть один и толькоодин прямой суперкласс (одиночное наследование).В отсутствие какого-либо другого явного суперкласса каждый класс неявно является подклассом Object.

Source https://docs.oracle.com/javase/tutorial/java/IandI/subclasses.html