java - 如何使用hibernate在spring的实体类中获取inet

Question

public List getAllEmployees() 
{
    return sessionFactory.openSession().createSQLQuery("select * from employee order by eid").list();
}

employee 表有一列ipadres，类型是postgresql 中的inet。

@Entity
@Table(name="employee")
public class Employee {
    @Id
    @Column(name="eid")
    private int eid;
    private int dept_id;
    private String name;
    private String address;
    private String project;
    private String password;
    @Column(name="ipadres")
    private String ipadres;
    private double salary;
    private Date Doj;

这是我的 pojo 类(class)。我已将 ipadres 作为字符串，但它给了我以下异常。

org.hibernate.MappingException: No Dialect mapping for JDBC type: 1111

Answer 1

当我必须创建自定义货币类型时，我遇到了类似的问题。

解决方案是创建您自己的实现UserType 接口(interface)的类。

这是一篇关于此事的精彩文章:example

简而言之，您有兴趣实现以下方法:

import org.hibernate.usertype.UserType;

public InetType impelements UserType{

public Class<String> returnedClass() {
    return String.class;
}

public int[] sqlTypes() {
    return new int[] { Types.OTHER }; // as the db type is inet and not directly transmutable to hibernate type.
}

public Object nullSafeGet(ResultSet resultSet, String[] names, Object owner)
        throws HibernateException, SQLException {
    String value = (String) Hibernate.STRING.nullSafeGet(resultSet, names[0]);
    return value;        
}

public void nullSafeSet(PreparedStatement preparedStatement, Object value, int index)
        throws HibernateException, SQLException {
    Hibernate.STRING.nullSafeSet(preparedStatement, 
            (value != null) ? value.toString() : null, index);
}
}

然后在你的Employee实体中，你需要添加类型定义注解:

@Entity
@Table(name="employee")
@TypeDefs(value={@TypeDef(name="inetType",typeClass=InetType.class)})
public class Employee {

    @Column(name="ipadres")
    @Type(type="inetType")
    private String ipadres;