休止状態で条件付きの結合と副問合せを使用する

Question

ロットを検索しました。しかし、私の場合は解決策を見つけることができません。私は次のクエリのための休止状態の基準を作成したい。

SELECT * FROM patient as p1 LEFT OUTER JOIN (SELECT * FROM patient_caller_admin_map WHERE caller_admin_id='1') as pca ON p1.patient_id=pca.patient_id; 

私はDetachedCriteriaを、基準を経て、次のものを作成しました。しかし、両方を結合することによってLEFT_JOINを使用する方法を知らないでください。

DetachedCriteria inner=DetachedCriteria.forClass(PatientCallerAdminMap.class, "patientCallerAdmin"); 
Criteria cr1=this.sessionFactory.getCurrentSession().createCriteria(Patient.class,"patient"); 

PatientCallerAdminMapエンティティ：

/** 
* PatientCallerAdminMap generated by hbm2java 
*/ 
@Entity 
@Table(name = "patient_caller_admin_map", catalog = "test") 
public class PatientCallerAdminMap implements java.io.Serializable { 

private PatientCallerAdminMapId id; 
private CallerAdmin callerAdmin; 
private Caller caller; 
private Patient patient; 
private String notes; 
private Integer isArchived; 
private Integer patientStatus; 
private Set<CallLog> callLogs = new HashSet<CallLog>(0); 
private Set<CallLog> callLogs_1 = new HashSet<CallLog>(0); 

public PatientCallerAdminMap() { 
} 

public PatientCallerAdminMap(PatientCallerAdminMapId id, 
     CallerAdmin callerAdmin, Patient patient) { 
    this.id = id; 
    this.callerAdmin = callerAdmin; 
    this.patient = patient; 
} 

public PatientCallerAdminMap(PatientCallerAdminMapId id, 
     CallerAdmin callerAdmin, Caller caller, Patient patient, 
     String notes, Integer isArchived, Integer patientStatus, 
     Set<CallLog> callLogs, Set<CallLog> callLogs_1) { 
    this.id = id; 
    this.callerAdmin = callerAdmin; 
    this.caller = caller; 
    this.patient = patient; 
    this.notes = notes; 
    this.isArchived = isArchived; 
    this.patientStatus = patientStatus; 
    this.callLogs = callLogs; 
    this.callLogs_1 = callLogs_1; 
} 

@EmbeddedId 
@AttributeOverrides({ 
     @AttributeOverride(name = "patientId", column = @Column(name = "patient_id", nullable = false)), 
     @AttributeOverride(name = "callerAdminId", column = @Column(name = "caller_admin_id", nullable = false)) }) 
public PatientCallerAdminMapId getId() { 
    return this.id; 
} 

public void setId(PatientCallerAdminMapId id) { 
    this.id = id; 
} 

@ManyToOne(fetch = FetchType.LAZY) 
@JoinColumn(name = "caller_admin_id", nullable = false, insertable = false, updatable = false) 
public CallerAdmin getCallerAdmin() { 
    return this.callerAdmin; 
} 

public void setCallerAdmin(CallerAdmin callerAdmin) { 
    this.callerAdmin = callerAdmin; 
} 

@ManyToOne(fetch = FetchType.LAZY) 
@JoinColumn(name = "caller_id") 
public Caller getCaller() { 
    return this.caller; 
} 

public void setCaller(Caller caller) { 
    this.caller = caller; 
} 

@ManyToOne(fetch = FetchType.LAZY) 
@JoinColumn(name = "patient_id", nullable = false, insertable = false, updatable = false) 
public Patient getPatient() { 
    return this.patient; 
} 

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

@Column(name = "notes", length = 600) 
public String getNotes() { 
    return this.notes; 
} 

public void setNotes(String notes) { 
    this.notes = notes; 
} 

@Column(name = "is_archived") 
public Integer getIsArchived() { 
    return this.isArchived; 
} 

public void setIsArchived(Integer isArchived) { 
    this.isArchived = isArchived; 
} 

@Column(name = "patient_status") 
public Integer getPatientStatus() { 
    return this.patientStatus; 
} 

public void setPatientStatus(Integer patientStatus) { 
    this.patientStatus = patientStatus; 
} 

@OneToMany(fetch = FetchType.LAZY, mappedBy = "patientCallerAdminMap") 
public Set<CallLog> getCallLogs() { 
    return this.callLogs; 
} 

public void setCallLogs(Set<CallLog> callLogs) { 
    this.callLogs = callLogs; 
} 

@OneToMany(fetch = FetchType.LAZY, mappedBy = "patientCallerAdminMap") 
public Set<CallLog> getCallLogs_1() { 
    return this.callLogs_1; 
} 

public void setCallLogs_1(Set<CallLog> callLogs_1) { 
    this.callLogs_1 = callLogs_1; 
} 

} 

患者エンティティクラス：

@Entity 
@Table(name = "patient", catalog = "test") 
public class Patient implements java.io.Serializable { 

    private String patientId; 
    private String addedDate; 
    private String name; 
    private String dateOfBirth; 
    private String gender; 
    private String address; 
    private String phoneNumber; 
    private Integer tier; 
    private Integer patientStatus; 
    private Integer status; 
    private Set<PatientCallerAdminMap> patientCallerAdminMaps = new HashSet<PatientCallerAdminMap>(
      0); 

    public Patient() { 
    } 

    public Patient(String patientId) { 
     this.patientId = patientId; 
    } 

    public Patient(String patientId,String addedDate, String timeOfCrash, 
      String name, String dateOfBirth, String gender, 
      String address, 
      String phoneNumber,Integer tier, Integer patientStatus, 
      Integer status, 
      Set<PatientCallerAdminMap> patientCallerAdminMaps, 
      ) { 
     this.patientId = patientId; 
     this.addedDate = addedDate; 
     this.name = name; 
     this.dateOfBirth = dateOfBirth; 
     this.gender = gender; 
     this.address = address; 
     this.phoneNumber = phoneNumber; 
     this.tier=tier; 
     this.patientStatus = patientStatus; 
     this.status = status; 
     this.patientCallerAdminMaps = patientCallerAdminMaps; 
    } 

    @Id 
    @Column(name = "patient_id", unique = true, nullable = false) 
    public String getPatientId() { 
     return this.patientId; 
    } 

    public void setPatientId(String patientId) { 
     this.patientId = patientId; 
    } 
    @Column(name = "added_date", length = 45) 
    public String getAddedDate() { 
     return addedDate; 
    } 

    public void setAddedDate(String addedDate) { 
     this.addedDate = addedDate; 
    } 

    @Column(name = "name", length = 100) 
    public String getName() { 
     return this.name; 
    } 

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

    @Column(name = "date_of_birth", length = 45) 
    public String getDateOfBirth() { 
     return this.dateOfBirth; 
    } 

    public void setDateOfBirth(String dateOfBirth) { 
     this.dateOfBirth = dateOfBirth; 
    } 

    @Column(name = "gender", length = 5) 
    public String getGender() { 
     return this.gender; 
    } 

    public void setGender(String gender) { 
     this.gender = gender; 
    } 

    @Column(name = "address", length = 200) 
    public String getAddress() { 
     return this.address; 
    } 

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

    @Column(name = "phone_number", length = 20) 
    public String getPhoneNumber() { 
     return this.phoneNumber; 
    } 

    public void setPhoneNumber(String phoneNumber) { 
     this.phoneNumber = phoneNumber; 
    } 


    @Column(name = "tier") 
    public Integer getTier() { 
     return this.tier; 
    } 

    public void setTier(Integer tier) { 
     this.tier = tier; 
    } 

    @Column(name = "patient_status") 
    public Integer getPatientStatus() { 
     return this.patientStatus; 
    } 

    public void setPatientStatus(Integer patientStatus) { 
     this.patientStatus = patientStatus; 
    } 

    @Column(name = "status") 
    public Integer getStatus() { 
     return this.status; 
    } 

    public void setStatus(Integer status) { 
     this.status = status; 
    } 

    @OneToMany(fetch = FetchType.LAZY, mappedBy = "patient") 
    public Set<PatientCallerAdminMap> getPatientCallerAdminMaps() { 
     return this.patientCallerAdminMaps; 
    } 

    public void setPatientCallerAdminMaps(
      Set<PatientCallerAdminMap> patientCallerAdminMaps) { 
     this.patientCallerAdminMaps = patientCallerAdminMaps; 
    } 


} 

これを解決するために助けてください。

Answer 1

クエリが簡単になるので、たぶん、あなたは、サブクエリを使用せずにこれを達成することができます：あなたが基準にそれを行うことはできません「*選択」については

Criteria cr1=this.sessionFactory.getCurrentSession().createCriteria(Patient.class,"patient"); 
cr2=cr1.createCriteria("patientCallerAdminMaps ",CriteriaSpecification.LEFT_JOIN); 
cr3= cr2.createCriteria("callerAdmin",CriteriaSpecification.LEFT_JOIN); 
cr3.add(Restrictions.eq("id", "1")); 

を。この基準は、患者エンティティのリストを返す。

本当にしたい場合*あなたはsubcriteria上の別名を追加し、明示的に