从Hibernate中的外键表获取主键表的结果

Question

我刚开始冬眠并尝试使用标准。我一直在从2个表（即主外键在realtion中的表）中获取结果。

我有Carpooler和SourceToDestination细节DTO，现在基于用户搜索数据，我想填充Carpooler Object，其中包含SourceToDestination细节，但我没有得到它，不知道如何使用标准API。

public class Carpooler implements Serializable{

    private long carpoolerId;
    private String drivingLicenceNumber=null;
    private String userType=null;![enter image description here][1]
    private User user=null;
    private List<VehicleDetails> listOfVehicleDetails=null;
    private List<SourceToDestinationDetails> listOfSourceToDestinationDetails=null;
    private Date carpoolerCreationDate;
}

public class SourceToDestinationDetails implements Serializable{

    private static final long serialVersionUID = -7158985673279885525L;

    private long sourceToDestinationId;
    private String sourcePlace=null;
    private String destinationPlace=null;
    private String inBetweenPlaces=null;

    private String sourceLeavingTime=null;
}

Criteria criteria1 = getSession().createCriteria(SourceToDestinationDetails.class);
criteria1.add(Restrictions.like("sourcePlace", "%" + from + "%"));
criteria1.add(Restrictions.like("destinationPlace", "%" + to + "%"));
List<SourceToDestinationDetails> listOfExactMatchCarpooler = criteria1.list();  

通过以上标准API，我只得到了SourceToDestination细节DTO记录，但是现在我也需要拼车记录，我不知道如何在SourceToDestination细节表中获得匹配Carpooler_id的拼车记录。

我是说如果用户给出，

String from = "Bellandur";
    String to = "Silk Board";

然后结果应该是List




Answer 1

可以通过注释来实现。您可以在SourceToDestinationDetails类中使用@OneToMany注释，如下所示：，

public class SourceToDestinationDetails implements Serializable{

    private static final long serialVersionUID = -7158985673279885525L;
    @Column
    private long sourceToDestinationId;
    @Column
    private String sourcePlace=null;
    @Column
    private String destinationPlace=null;
    @Column
    private String inBetweenPlaces=null;
    @Column
    private String sourceLeavingTime=null;

    @OneToMany(mappedBy = "carpooler_id", cascade = CascadeType.ALL)
    private Set<Carpooler> carpoolers;
}

如果您想使用HIBERNATE XML实现相同的目标，请按以下方式声明XML

  <set name="carpoolers" table="source_destination" 
            inverse="true" lazy="true" fetch="select">
        <key>
            <column name="carpooler_id" not-null="true" />
        </key>
        <one-to-many class="com.test.Carpooler" />
    </set>

在这种情况下，您的模型类将

public class SourceToDestinationDetails implements Serializable{

    private static final long serialVersionUID = -7158985673279885525L;

    private long sourceToDestinationId;
    private String sourcePlace=null;
    private String destinationPlace=null;
    private String inBetweenPlaces=null;

    private String sourceLeavingTime=null;
    private Set<StockDailyRecord> carpoolers = 
                new HashSet<StockDailyRecord>();
}

我通常更喜欢注释，而不是丑陋的XML