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hibernate: Unable to locate appropriate constructor on class - HQL

When I trying to execute this HQL to return an object Ponto I receive this error:

ERROR [org.hibernate.hql.PARSER] (http-localhost-127.0.0.1-8080-2) Unable to locate appropriate constructor on class [br.com.cdv.model.entity.Ponto] [cause=org.hibernate.PropertyNotFoundException: no appropriate constructor in class: br.com.cdv.model.entity.Ponto]

DAO

    @SuppressWarnings("unchecked")
    @Override
    public List<Ponto> listLoja(Integer idLoja) {

        Query q = getSession().createQuery("select new Ponto(0,ss.cliente,ss.loja,null,null,null,null,null,sum(qtdPontos),'',0) "
            + "from Ponto as ss where ss.loja.id = :idLoja "
            + "group by ss.cliente, ss.loja");  

        q.setParameter("idLoja", idLoja);

        return (List<Ponto>) q.list();
    }  

My Entity / Class

@Entity
@Table (name = "ponto")
public class Ponto implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue
    private Integer id;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name="cliente", nullable=true)
    private UsuarioCliente cliente;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name="loja", nullable=false)
    private UsuarioLoja loja;

    @Column(name="dataCriacao")
    private Date dataCriacao;

    @Column(name="dataUtilizado", length=12, nullable=true)
    private Date dataUtilizado;

    @Column(name="dataExpira")
    private Date dataExpira;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "funcionario", nullable=true)
    private Funcionario funcionario;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "pontoReceber", nullable=true)
    private PontoReceber pontoReceber;

    @Column(name="qtdPontos", nullable=false)
    private long qtdPontos;

    @Column(name="obsPontos", nullable = true,length=300)
    private String obsPontos;

    @NotEmpty
    @Column(name="tipo",nullable = false)
    private Integer tipo;

    public Ponto(Integer id, UsuarioCliente cliente,UsuarioLoja loja, Date dataCriacao, Date dataUtilizado,
              Date dataExpira, Funcionario funcionario, PontoReceber pontoReceber, long qtdPontos, String obsPontos, Integer tipo) {
        setId(id);
        setCliente(cliente);
        setLoja(loja);
        setDataCriacao(dataCriacao);
        setDataUtilizado(dataUtilizado);
        setDataExpira(dataExpira);
        setFuncionario(funcionario);
        setPontoReceber(pontoReceber);
        setQtdPontos(qtdPontos);
        setObsPontos(obsPontos);
        setTipo(tipo);
    }
        // getters and setters
    }

Control:

@RequestMapping("/listarClientes")
    public String listarClientesPontos(Map<String, Object> map, HttpSession session) {

        ...

        List<Ponto> pontos = pontoService.listLoja(loja.getId());

        map.put("pontos", pontos);

        return "listaClientesPonto";
    }
}

View:

<%@taglib uri="http://java.sun.com/jsp/jstl/core" prefix="c"%>
<%@taglib uri="http://www.springframework.org/tags/form" prefix="form"%>
<%@taglib uri="http://www.springframework.org/tags" prefix="spring"%>


    <body>
        <h2>Lista Clientes</h2>

        <table>
          <tr>
            <th>ID Cliente</th>
            <th>CPF Cliente</th>
            <th>Pontos totais:</th>
          </tr>

          <c:forEach items="${pontos}" var="ponto" varStatus="count">
          <tr>
            <td>${ponto.cliente.id}</td>
            <td>${ponto.cliente.cpf}</td>
            <td>${ponto.qtdPontos}</td>
          </tr>
          </c:forEach>
        </table>
    </body>

Why I am getting this error?
Has any better why to receive this Object Ponto in a List?

obs.. without new Ponto(...) returns a list of Ponto with unidentified Objects[]

like image 890
Lipo Avatar asked Mar 24 '15 02:03

Lipo


2 Answers

Check these things:

1- If you make a constructor with parameters; you should provide the constructor with no parameters, explicity;

2- Make sure your ID entity is int/Integer;

3- Make your Entity java.io.Serializable by implementing;

4- Make your parameter-less (default) constructor public or default access modifier;

like image 174
G Bisconcini Avatar answered Sep 18 '22 16:09

G Bisconcini


Found the problem... I made some bad constructors, so I edited the constructors in my entity:

@Entity
@Table (name = "ponto")
public class Ponto implements java.io.Serializable {

    @Id
    @GeneratedValue
    private Integer id;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name="cliente", nullable=true)
    private UsuarioCliente cliente;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name="loja", nullable=false)
    private UsuarioLoja loja;

    @Column(name="dataCriacao")
    private Date dataCriacao;

    @Column(name="dataUtilizado", length=12, nullable=true)
    private Date dataUtilizado;

    @Column(name="dataExpira")
    private Date dataExpira;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "funcionario", nullable=true)
    private Funcionario funcionario;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "pontoReceber", nullable=true)
    private PontoReceber pontoReceber;

    @Column(name="qtdPontos", nullable=false)
    private long qtdPontos;

    @Column(name="obsPontos", nullable = true,length=300)
    private String obsPontos;

    @NotEmpty
    @Column(name="tipo",nullable = true)
    private Integer tipo;

    public Ponto() {
    }

    public Ponto(UsuarioCliente cliente, UsuarioLoja loja, long qtdPontos) {
        this.cliente = cliente;
        this.loja = loja;
        this.qtdPontos = qtdPontos;
    }
    // getters and setters
}

and HQL:

    Query q = getSession().createQuery("select new Ponto(ss.cliente,ss.loja,sum(ss.qtdPontos) as qtdPontos) "
            + "from Ponto as ss where ss.loja.id = :idLoja "
            + "group by ss.cliente, ss.loja");
    q.setParameter("idLoja", idLoja);

I am crying like a baby, four days with this issue.

Thanks for the directions Thufir Hawat.

like image 40
Lipo Avatar answered Sep 18 '22 16:09

Lipo