API criterios No se puede localizar el constructor apropiado en clase

Jacob:

Cuando yo estoy poniendo en práctica la API de criterios de unión para mi estudio de arranque de primavera, he intentado unirse a las clases 2 y ir a buscar el resultado. Pero cuando me estoy poniendo en práctica y ejecutar estoy consiguiendo el error siguiente,

Unable to locate appropriate constructor on class [com.spacestudy.model.Investigator]. Expected arguments are: com.spacestudy.model.Employee
[cause=org.hibernate.PropertyNotFoundException: no appropriate constructor in class: com.spacestudy.model.Investigator]

Y mi clase Employee.java como la siguiente,

@Entity
@Table(name="employee")
public class Employee implements Serializable
{
private static final long serialVersionUID = 1L;

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY, generator = "employee_seq_generator")
@SequenceGenerator(name = "employee_seq_generator", sequenceName = "employee_seq",allocationSize=1)
@Column(name="nemp_id",columnDefinition="serial")
public Integer nEmpId;

@Column(name="semp_name")
public String sEmpName;

@Column(name="sdesignation")
public String sDesignation;

@Column(name="ninst_id")
public Integer nInstId;

@Column(name="ndept_id")
public Integer nDeptId;

@Column(name="sclient_emp_id")
public String sClientEmpId;

@Column(name="ntemp_emp_id")
public Integer nTempEmpId;

@Column(name="bis_paid")
public boolean bIsPaid=true;

@Column(name="sunpaid_comment")
public String sUnpaidComment;

@ManyToOne(optional=true)
@JoinColumn(name="ndept_id", insertable = false, updatable = false)
public Department department;

@OneToMany(cascade = CascadeType.ALL,mappedBy="nEmpId")
public Set<Investigator> employeeInvestigatorJoinMapping;

public Employee() {
}

public Employee(Integer nEmpId, String sEmpName, String sDesignation, Integer nInstId, Integer nDeptId,
        String sClientEmpId, Integer nTempEmpId, boolean bIsPaid, String sUnpaidComment, Department department,
        Set<Investigator> employeeInvestigatorJoinMapping) {
    super();
    this.nEmpId = nEmpId;
    this.sEmpName = sEmpName;
    this.sDesignation = sDesignation;
    this.nInstId = nInstId;
    this.nDeptId = nDeptId;
    this.sClientEmpId = sClientEmpId;
    this.nTempEmpId = nTempEmpId;
    this.bIsPaid = bIsPaid;
    this.sUnpaidComment = sUnpaidComment;
    this.department = department;
    this.employeeInvestigatorJoinMapping = employeeInvestigatorJoinMapping;
   }    
 }

Y mi segunda clase Investigator.java,

@Entity
@Table(name = "investigator")
@JsonInclude(JsonInclude.Include.NON_NULL) // avoiding null values
public class Investigator implements Serializable 
{
private static final long serialVersionUID = 1L;

@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "investigator_seq_generator")
@SequenceGenerator(name = "investigator_seq_generator", sequenceName = "investigator_seq")

@Column(name="ninvestigator_id")
public Integer nInvestigatorId;

@Column(name="sinvestigator_name")
public String sInvestigatorName;

@Column(name="ninst_id")
public Integer nInstId;

@Column(name="stitle")
public String sTitle;

@Column(name="ntemp_investigator_id")
public Integer nTempInvestigatorId;

@ManyToOne(optional = false)
@JoinColumn(name="nemp_id",referencedColumnName="nemp_id")
public Employee nEmpId;

// Default Constructor.
public Investigator()
{
}
public Investigator(Integer nInvestigatorId, String sInvestigatorName, Integer nInstId, String sTitle,
        Integer nTempInvestigatorId, Employee nEmpId) {
    super();
    this.nInvestigatorId = nInvestigatorId;
    this.sInvestigatorName = sInvestigatorName;
    this.nInstId = nInstId;
    this.sTitle = sTitle;
    this.nTempInvestigatorId = nTempInvestigatorId;
    this.nEmpId = nEmpId;
   }
}

E implementado la API Criterios de unirse como la siguiente,

CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Investigator> cq = cb.createQuery(Investigator.class);
Root<Employee> rootInvestigator = cq.from(Employee.class);
Join<Employee ,Investigator> resultEmployeeMappingObj 
    = rootInvestigator.join("employeeInvestigatorJoinMapping");
cq.multiselect(rootInvestigator);
cq.where(cb.equal(resultEmployeeMappingObj.get("nEmpId"), 21638));
List<Investigator> results = em.createQuery(cq).getResultList();
return results;

¿Qué hice mal?

Evgeniy Khyst :

API criterios

Usted tiene algunos errores en la consulta de la API Criteria.

Los que trabajan uno es similar al siguiente

@Transactional(readOnly = true)
public List<Investigator> findByEmployeeId(int employeeId) {
  CriteriaBuilder criteriaBuilder = em.getCriteriaBuilder();
  CriteriaQuery<Investigator> query = criteriaBuilder.createQuery(Investigator.class);
  Root<Investigator> investigator = query.from(Investigator.class);
  Join<Investigator, Employee> employees = investigator.join("nEmpId");
  query.select(investigator)
      .where(criteriaBuilder.equal(employees.get("nEmpId"), employeeId));
  TypedQuery<Investigator> typedQuery = em.createQuery(query);
  List<Investigator> investigators = typedQuery.getResultList();
  log.debug("Investigators: {}", investigators);
  return investigators;
}

Primavera de datos JPA

Además, si su aplicación se basa en la infraestructura Spring después de cambiar el nombre de un par de campos que puede utilizar la primavera de datos JPA y hacer consulta no escribir en absoluto.

Employee entidad:

@Entity
@Table(name = "employee")
public class Employee implements Serializable {

  @Id
  @GeneratedValue(strategy = GenerationType.IDENTITY, generator = "employee_seq_generator")
  @SequenceGenerator(name = "employee_seq_generator", sequenceName = "employee_seq", allocationSize = 1)
  @Column(name = "nemp_id", columnDefinition = "serial")
  public Integer id;

  //...

  @OneToMany(cascade = CascadeType.ALL, mappedBy = "employee")
  public Set<Investigator> investigators = new HashSet<>();

  //...
}

Investigator entidad:

@Entity
@Table(name = "investigator")
@JsonInclude(JsonInclude.Include.NON_NULL) // avoiding null values
public class Investigator implements Serializable {

  //...

  @ManyToOne(optional = false)
  @JoinColumn(name = "nemp_id", referencedColumnName = "nemp_id")
  public Employee employee;

  //...
}

Primavera de datos JPA interfaz de repositorio:

public interface InvestigatorRepository extends JpaRepository<Investigator, Integer> {

  List<Investigator> findByEmployeeId(int employeeId);
}

Eso es. Ahora simplemente puede inyectar el repositorio y usarlo:

@Autowired
private InvestigatorRepository investigatorRepository;

public void testQuery() {
    investigatorRepository.findByEmployeeId(employee.getId()));
}