Hibernate hql Internal join Eanger fetch对一对多进行抓取，获取冗余的父对象

Question

我有如下实体

课程进展

@Entity
public class ProgressNote implements Serializable{

    @Id
    private NotesKey notesKey = new NotesKey();

    private Set<PatientObjective> patientObjectives;

    public NotesKey getNotesKey() {
        return notesKey;
    }
    public void setNotesKey(NotesKey notesKey) {
        this.notesKey = notesKey;
    }


    @OneToMany(fetch=FetchType.LAZY)
    @Access(AccessType.PROPERTY)
    @JoinColumns({
        @JoinColumn(name="noteNumber",referencedColumnName="noteNumber"),
        @JoinColumn(name="ddate",referencedColumnName="ddate"),
        @JoinColumn(name="patient_id",referencedColumnName="patient_id")
    })
    public Set<PatientObjective> getPatientObjectives() {
        return patientObjectives;
    }
    public void setPatientObjectives(Set<PatientObjective> patientObjectives) {
        this.patientObjectives = patientObjectives;
    }


}

课堂笔记

@Embeddable
public class NotesKey implements Serializable{

    private Byte noteNumber;

    @Temporal(javax.persistence.TemporalType.DATE)
    @Column(name="ddate")
    private Date noteDate;

    private Patient patient;

    public Byte getNoteNumber() {
        return noteNumber;
    }

    public void setNoteNumber(Byte noteNumber) {
        this.noteNumber = noteNumber;
    }

    public Date getNoteDate() {
        return noteDate;
    }

    public void setNoteDate(Date noteDate) {
        this.noteDate = noteDate;
    }

    @ManyToOne(fetch=FetchType.LAZY)
    @Access(AccessType.PROPERTY)
    public Patient getPatient() {
        return patient;
    }

    public void setPatient(Patient patient) {
        this.patient = patient;
    }

    @Override
    public int hashCode() {
        ........
    }

    @Override
    public boolean equals(Object obj) {
        ........
    }

}

目标类

@Entity
public class PatientObjective implements Serializable{

    @Id
    private PatientObjectiveKey patientObjectiveKey;


    public PatientObjectiveKey getPatientObjectiveKey() {
        return patientObjectiveKey;
    }
    public void setPatientObjectiveKey(PatientObjectiveKey patientObjectiveKey) {
        this.patientObjectiveKey = patientObjectiveKey;
    }

}

类PatientObjectiveKey

@Embeddable
public class PatientObjectiveKey implements Serializable{

    private Objective objective;

    private Byte noteNumber;

    @Temporal(javax.persistence.TemporalType.DATE)
    @Column(name="ddate")
    private Date noteDate;

    private Patient patient;

    @ManyToOne(fetch=FetchType.LAZY)
    @Access(AccessType.PROPERTY)
    public Objective getObjective() {
        return objective;
    }

    public void setObjective(Objective objective) {
        this.objective = objective;
    }

    public Byte getNoteNumber() {
        return noteNumber;
    }

    public void setNoteNumber(Byte noteNumber) {
        this.noteNumber = noteNumber;
    }

    public Date getNoteDate() {
        return noteDate;
    }

    public void setNoteDate(Date noteDate) {
        this.noteDate = noteDate;
    }

    @ManyToOne(fetch=FetchType.LAZY)
    @Access(AccessType.PROPERTY)
    public Patient getPatient() {
        return patient;
    }

    public void setPatient(Patient patient) {
        this.patient = patient;
    }

    @Override
    public int hashCode() {
        .......
    }

    @Override
    public boolean equals(Object obj) {
        ........
    }

}

在我使用这个命名查询之后

@namedQuery（name="findAllProgressNoteAnd Objals"， query="from ProgressNote p内连接获取p.patientObjals as o其中p.notesKey.patient.id=： ResidentId）"）使用SpringHibernate模板findByNamedQueryAndNamedParam

在表I中有两个ProgressNote，一个只有一个目标，另一个有17个目标。我在Java中得到如下结果。

查询生成为

select progressno0_.ddate as ddate46_0_, progressno0_.noteNumber as noteNumber46_0_, progressno0_.patient_id as patient17_46_0_, patientobj1_.ddate as ddate36_1_, patientobj1_.noteNumber as noteNumber36_1_, patientobj1_.objective_id as objective5_36_1_, patientobj1_.patient_id as patient6_36_1_, progressno0_.assessment as assessment46_0_, progressno0_.bloodPressure1 as bloodPre4_46_0_, progressno0_.bloodPressure2 as bloodPre5_46_0_, progressno0_.creationDate as creation6_46_0_, progressno0_.height as height46_0_, progressno0_.lastUpdatedDate as lastUpda8_46_0_, progressno0_.plans as plans46_0_, progressno0_.status as status46_0_, progressno0_.subject as subject46_0_, progressno0_.temprature as temprature46_0_, progressno0_.tempratureUnit as temprat13_46_0_, progressno0_.lastid as lastid46_0_, progressno0_.waist as waist46_0_, progressno0_.weight as weight46_0_, progressno0_.weightUnit as weightUnit46_0_, patientobj1_.remark as remark36_1_, patientobj1_.value as value36_1_, patientobj1_.ddate as ddate46_0__, patientobj1_.noteNumber as noteNumber46_0__, patientobj1_.patient_id as patient6_46_0__, patientobj1_.ddate as ddate0__, patientobj1_.noteNumber as noteNumber0__, patientobj1_.objective_id as objective5_0__, patientobj1_.patient_id as patient6_0__ from pnheader progressno0_ inner join pnobjremark patientobj1_ on progressno0_.ddate=patientobj1_.ddate and progressno0_.noteNumber=patientobj1_.noteNumber and progressno0_.patient_id=patientobj1_.patient_id where progressno0_.patient_id=? 

问题

我看到的问题是为什么我会得到ProgressNote的多个单一实例（id=152）。有没有办法避免这种重复？

如果有人向我解释或指出正确的方向，我将不胜感激。

使用Hibernate 3.6、Mysql 5.6.10

Answer 1

使用不同的：

select distinct p from ProgressNote p 
inner join fetch p.patientObjectives as o 
where p.notesKey.patient.id = :patientId)