Question

当我尝试构建项目时，我遇到了@Autowired和我的JpaReposytory类的问题，问题出现在这里。

这是我的Jpa Repository类。

public interface IJpaRepositoryLivrariaDAO extends JpaRepository<EntidadeDominio, Long> {
    List<Cliente> findByCliente(Cliente cliente); // whem I put this line of code I've get the exeption below                                           
}

DAO实施

@Repository
public class ClienteDAO implements IDAO {

    @Autowired
    List<Cliente> cliente;

    @Autowired
    IJpaRepositoryLivrariaDAO jpaRepositoryLivraria;


    @Override
    public void consultar(ModelMap modelMap) {

        Cliente c = (Cliente) modelMap.get("object");

        modelMap.addAttribute("object", jpaRepositoryLivraria
            .findByCliente(c));

}

例外

2017-10-10 21:18:58.524 WARN 4000 --- [restartedMain] ationConfigEmbeddedWebApplicationContext：在上下文初始化期间遇到异常 - 取消刷新尝试：

org.springframework.beans.factory.UnsatisfiedDependencyException：创建名为'alterarCommand'的bean时出错：通过字段'fachada'表示的不满意的依赖关系;

嵌套异常是org.springframework.beans.factory.BeanCreationException：创建名为'fachada'的bean时出错：资源依赖注入失败;

嵌套异常是org.springframework.beans.factory.UnsatisfiedDependencyException：创建名为'clienteDAO'的bean时出错：通过字段'jpaRepositoryLivraria'表示的不满意的依赖关系;

嵌套异常是org.springframework.beans.factory.BeanCreationException：创建名为'IJpaRepositoryLivrariaDAO'的bean时出错：init方法的调用失败;

嵌套异常是org.springframework.data.mapping.PropertyReferenceException：找不到类型为Cliente的属性cliente！

如果我错了，请纠正我，问题是我的客户列表中没有客户端或类似的东西

列出findByCliente（Cliente cliente）;

提前感谢您的帮助。

更新1。

好吧，基于我做出一些改动的答案，试图使代码有效。

首先，如果我在EntidadeDominio中放置一个属性Cliente，它会弄乱我的数据库，所以我创建了一个新的JPARepository，试图让代码更容易理解和执行

public interface IJpaRepositoryClienteDAO extends JpaRepository<Cliente, Long> {

    // I don't know the best way to use @Query, what I'm trying to do here
    // is to select * from Cliente where any of the clients 
    // atributes matches with their corresponding columns

    @Query("select c from Cliente c where c like %?1")
    List<Cliente> findByCliente(Cliente cliente);

}

然后在

中调用此方法

public class ClienteDAO 

    @Autowired
    IJpaRepositoryClienteDAO jpaRepositoryCliente;

    public void consultar(ModelMap modelMap) {

        Cliente c = (Cliente) modelMap.get("object");

        modelMap.remove("object");
        modelMap.remove("cliente");

        modelMap.addAttribute("object", jpaRepositoryLivraria.findByCliente(c));
    }
}

这是我的Cliente课程和他的绑定课程，以便更好地保留

@Entity
@Component
@DiscriminatorValue("Cliente")
public class Cliente extends EntidadeDominio {

    @Embedded
    private Genero genero;
    private boolean ativo;

    @OneToMany(mappedBy = "cliente", targetEntity = CartaoCredito.class, cascade = CascadeType.ALL, orphanRemoval = true)
    private List<CartaoCredito> cartaoCredito = new ArrayList<>();

    @OneToMany(mappedBy = "cliente", targetEntity = Endereco.class, cascade = CascadeType.ALL, orphanRemoval = true)
    private List<Endereco> endereco = new ArrayList<>();
}

@Entity
@Component
public class CartaoCredito {

    @Id
    @GeneratedValue( strategy=GenerationType.AUTO )
    private long id_cartao_credito;
    private String numero;
    private String nomeCartao;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "id_cliente")
    private Cliente cliente;
}

@Entity
@Component
public class Endereco {

    @Id
    @GeneratedValue
    private Long id_endereco;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "id_cliente")
    private Cliente cliente;
}

这样做会改变代码的工作原理，现在的问题是我得到了这个错误

java.lang.IllegalArgumentException：参数值[％com.les.livraria.dominio.cliente.Cliente@1a1147e]与预期类型[com.les.livraria.dominio.cliente.Cliente（n / a）]不匹配

正在我的ClienteDAO类中正确发送对象

modelMap.addAttribute("object", jpaRepositoryLivraria.findByCliente(c));

好吧，现在我要跟踪这个错误，看看源代码是什么问题，如果有人有任何关于如何处理这个问题的想法，我将非常感激，现在非常感谢你。

Answer 1

那里的事情错了......

1-异常告诉您spring不能找到cliente字段/属性，特别是：findByCliente(Cliente cliente)表示您需要在类Cliente cliente中指定的字段EntidadeDominio JpaRepository正在延长。

2-此外，由于JpaRepository属于EntidadeDominio，我希望findByCliente返回类型应为List<EntidadeDominio>而不是

我不确定你的目标是什么

Answer 2

我认为在您的EntidadeDominio类中你需要一个字段类型的Cliente然后你可以使用findByCliente

class EntidadeDominio  {

private Cliente  cliente  ;

// getter && setter 

}

并且你的dao findByCliente应该撤回EntidadeDominio List<EntidadeDominio>列表而不是cliente列表

public interface IJpaRepositoryLivrariaDAO extends 

    JpaRepository<EntidadeDominio, Long> {
        List<EntidadeDominio> findByCliente(Cliente cliente);  
    }

Answer 3

您IJpaRepositoryLivrariaDAO正在管理EntidadeDominio实体的持久性。

像这样更新您的存储库

public interface IJpaRepositoryLivrariaDAO extends JpaRepository<EntidadeDominio, Long> {

    List<EntidadeDominio> findByCliente(Cliente cliente); // This will return a list of EntidadeDominio

}

以后你可以这样做

List<EntidadeDominio> entidadeDominios =  repository.findByC...;
List<Cliente> clients = entidadeDominios.getCliente(); //Given that you have Cliente as composition in your EntidadeDominio entity.

Answer 4

谢谢@Zeromus，我认为可以从我的pojo类生成，现在很明显这是不可能的，我使用反射来进行我的通用查询，方法仍然没有完成，它需要异常治疗。我不知道这是否是解决这个问题的最佳解决方案，但我在这里发帖，也许对某人有帮助。

public class ReflectionQueryCreator {

    private String sql = "";
    private String nomeAtributo;
    private Object metodoValor;

    public String queryBuilderCliente(Cliente object, Class<Cliente> classe) {

        Method[] classMethods = classe.getMethods(); 
        Field[] classFields = classe.getDeclaredFields();

        for (Method method : classMethods) {
            try {
                if (method.getName().substring(0, 3).equals("get")) {

                    metodoValor = method.invoke(object);

                    for (Field field : classFields) {
                        nomeAtributo = field.getName();

                        if (method.getName().toLowerCase().endsWith(nomeAtributo.toLowerCase()))

                            if (!(metodoValor.equals("") || metodoValor.equals(0) || method.toString().contains("Id"))) {

                                if (method.getReturnType().getSimpleName().equals("String"))
                                    sql = sql + object.getClass().getSimpleName().toLowerCase() + "." + nomeAtributo + " like %" + metodoValor + "% and ";

                                else if (method.getReturnType().getSimpleName().equals("long") && metodoValor.equals(0))
                                    sql = sql + object.getClass().getSimpleName().toLowerCase() + "." + nomeAtributo + " = " + metodoValor + " and ";

                            } else
                                break;
                    }
                }
            } catch (IllegalAccessException ex) {ex.printStackTrace();}
              catch (IllegalArgumentException ex) {ex.printStackTrace();} 
              catch (InvocationTargetException ex) {ex.printStackTrace();}
        }
        return sql;
    }

@Autowired JPA Repository在春季启动时

4 个答案: