Java 休眠：无法在类上找到合适的构造函数 - HQL
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StackOverFlow
hibernate: Unable to locate appropriate constructor on class - HQL
提问by Lipo
When I trying to execute this HQL to return an object PontoI receive this error:
当我尝试执行此 HQL 以返回一个对象时，Ponto我收到此错误：
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]
错误 [org.hibernate.hql.PARSER] (http-localhost-127.0.0.1-8080-2) 无法在类 [br.com.cdv.model.entity.Ponto] 上找到适当的构造函数 [cause=org.hibernate。 PropertyNotFoundException：类中没有合适的构造函数：br.com.cdv.model.entity.Ponto]
DAO
道
@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
我的实体/班级
@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:
控制：
@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:
看法：
<%@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?
为什么我收到这个错误？
为什么在列表中接收这个 Object Ponto 有更好的理由吗？
obs.. without new Ponto(...)returns a list of Ponto with unidentified Objects[]
obs.. 不new Ponto(...)返回带有未识别对象的 Ponto 列表[]
采纳答案by Lipo
Found the problem... I made some bad constructors, so I edited the constructors in my entity:
发现问题...我做了一些不好的构造函数，所以我在我的实体中编辑了构造函数：
@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:
和 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.
我哭得像个婴儿，被这个问题困扰了四天。
Thanks for the directions Thufir Hawat.
感谢 Thufir Hawat 的指导。
回答by Thufir Hawat
Check these things:
检查这些事情：
1- If you make a constructor with parameters; you should provide the constructor with no parameters, explicity;
1- 如果您使用参数创建构造函数；您应该明确地提供不带参数的构造函数；
2- Make sure your ID entity is int/Integer;
2- 确保您的 ID 实体是 int/Integer；
3- Make your Entity java.io.Serializable by implementing;
3- 通过实现使您的实体 java.io.Serializable；
4- Make your parameter-less (default) constructor public or default access modifier;
4- 将您的无参数（默认）构造函数设为 public 或默认访问修饰符；
回答by Smart Coder
Making a public constructor with arguments may make it work.
使用参数创建公共构造函数可能会使其工作。
回答by gdrt
If you have made a selection (i.e. don't return all the columns of the table), make sure to have a constructor with the selected column(s) in your mapped class.
如果您进行了选择（即不返回表的所有列），请确保在您的映射类中有一个带有所选列的构造函数。
