如何使用Hibernate转换平面结果集

是否可以将SQL的结果映射到非平面对象？

List<Customer> customers = hibernateSession().createCriteria(CustomerDetailsView.class)
                .add(Restrictions.in("userName", userName))
                .setProjection(buildProjection())
                .setResultTransformer(Transformers.aliasToBean(Customer.class))
                .list();

在我的例子中,CustomerDetailsView具有扁平结构.但我需要将它映射到这样的对象：

public class Customer {
    private String userName;
    private String title;
    private String firstName;
    private String lastName;
    private String type;
    private String companyName;
    private AddressDetails addressDetails;
}

和

public class AddressDetails {
    private String countryCode;
    private String addressLine1;
    private String zipOrPostCode;
    private String city;
    private String countryDivisionName;
    private String countryDivisionCode;
    private String countryDivisionTypeCode;
    private String residentialAddress;
}

解决方法:

对的,这是可能的.您可以使用自定义变换器：
FluentHibernateResultTransformer.

您可以复制粘贴代码,或者通过Maven添加jar：fluent-hibernate-core.

您需要使用Criteria with Projections.请不要忘记指定投影别名(userName,addressDetails.countryCode)

Criteria criteria = session.createCriteria(Customer.class);
criteria.createAlias("addressDetails", "addressDetails", JoinType.LEFT_OUTER_JOIN);

criteria.setProjection(Projections.projectionList()
        .add(Projections.property("userName").as("userName"))
        .add(Projections.property("addressDetails.countryCode")
        .as("addressDetails.countryCode")));

List<Customer> customers = criteria.setResultTransformer(
        new FluentHibernateResultTransformer(Customer.class)).list();

与HQL一起使用

不可能将它与HQL一起使用,因为Hibernate不允许在HQL中嵌套别名

选择addressDetails.countryCode作为addressDetails.countryCode

addressDetails.countryCode别名将出错.

使用本机SQL

变换器可用于具有嵌套投影的本机SQL(与HQL相对).
在这种情况下,需要使用带引号的别名：

String sql = "select c.f_user_name as userName, d.f_country_code as \"addressDetails.countryCode\" "
        + "from customers c left outer join address_details d on c.fk_details = d.f_pid";

List<Customer> customers = session.createSQLQuery(sql)
        .setResultTransformer(new FluentHibernateResultTransformer(Customer.class))
        .list();

标签：java,hibernate,hibernate-criteria
来源： https://codeday.me/bug/20190918/1811199.html