JPA Query - 类错误中没有适当的构造函数答案

【问题标题】：JPA Query - no appropriate constructor in class errorJPA Query - 类错误中没有适当的构造函数
【发布时间】：2022-02-21 14:09:16
【问题描述】：

我有一个 Employee 实体类，它有很多列。我想从这个类中获取一些列，因此我使用了 dtos。我创建了一个新的 BaseEmployee 类并在 EmployeeRepository 中编写了查询。但是当我尝试运行应用程序时出现此错误：“类错误中没有适当的构造函数”。

@Entity
@Table(name = "employees")
@Data
@AllArgsConstructor
@NoArgsConstructor
@JsonAutoDetect(fieldVisibility = Visibility.ANY)
public class Employee {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "emp_no")
    private  int id;
    @Column(name = "birth_date")   
    private Date birthDate;
    @Column(name = "first_name")
    private String firstName;
    @Column(name = "last_name")
    private String lastName;
    @Column(name = "gender")
    private char gender;
    @Column(name = "hire_date")  
    private Date hireDate;

    @OneToMany(mappedBy = "employees")
    private List<Title> titles; 

    @OneToMany(mappedBy = "employees")
    private List<Salary> salary;

    @OneToMany(mappedBy = "employee")
    private List<DeptEmp> departmentList;

   @OneToMany(mappedBy = "employee")
   private List<DeptManager> managerDepartment;

}

我的 dto 类：

@Data
@NoArgsConstructor
@AllArgsConstructor
public class BaseEmployee {

    private  int id;
    private Date birthDate;
    private String firstName;
    private String lastName;
    private char gender;
    private Date hireDate;

}

Jpa 存储库：

public interface EmployeeRepository extends JpaRepository<Employee,Integer> {
    
    List<Employee> getByFirstNameContains(String firstName);
    List<Employee> getByFirstNameStartsWith(String firstName);


    @Query("Select new dev.serhat.employeeapi.models.dtos.BaseEmployee"
    + "(e.id, e.birthDate, e.firstName, e.lastName, e.gender, e.hireDate) From Employee e WHERE e.id = :id")
    Optional<BaseEmployee> getBaseEmployeeById(int id);
    
    
}

错误：

HH000400: Using dialect: org.hibernate.dialect.MySQL5InnoDBDialect
2022-02-20 16:33:42.533  INFO 24371 --- [  restartedMain] o.h.e.t.j.p.i.JtaPlatformInitiator       : HHH000490: Using JtaPlatform implementation: [org.hibernate.engine.transaction.jta.platform.internal.NoJtaPlatform]
2022-02-20 16:33:42.559  INFO 24371 --- [  restartedMain] j.LocalContainerEntityManagerFactoryBean : Initialized JPA EntityManagerFactory for persistence unit 'default'
2022-02-20 16:33:43.431 ERROR 24371 --- [  restartedMain] o.h.hql.internal.ast.ErrorTracker        :  Unable to locate appropriate constructor on class [dev.serhat.employeeapi.models.dtos.BaseEmployee]. Expected arguments are: int, java.util.Date, java.lang.String, java.lang.String, char, java.util.Date
[cause=org.hibernate.PropertyNotFoundException: no appropriate constructor in class: dev.serhat.employeeapi.models.dtos.BaseEmployee]
2022-02-20 16:33:43.443 ERROR 24371 --- [  restartedMain] o.h.hql.internal.ast.ErrorTracker        :  Unable to locate appropriate constructor on class [dev.serhat.employeeapi.models.dtos.BaseEmployee]. Expected arguments are: int, java.util.Date, java.lang.String, java.lang.String, char, java.util.Date
[cause=org.hibernate.PropertyNotFoundException: no appropriate constructor in class: dev.serhat.employeeapi.models.dtos.BaseEmployee]

org.hibernate.hql.internal.ast.DetailedSemanticException: Unable to locate appropriate constructor on class [dev.serhat.employeeapi.models.dtos.BaseEmployee]. Expected arguments are: int, java.util.Date, java.lang.String, java.lang.String, char, java.util.Date
        at org.hibernate.hql.internal.ast.tree.ConstructorNode.resolveConstructor(ConstructorNode.java:182) ~[hibernate-core-5.6.4.Final.jar:5.6.4.Final]

【问题讨论】：

标签： java spring-boot hibernate spring-data-jpa dto

【解决方案1】：

Lombok 生成的全参数构造函数调用java.util.Date 类型的两个参数。

当 hibernate 执行你的 jpql 时，它在构造函数中使用 org.hibernate.type.TimestampType 对象而不是 java.util.Date 对象。当然没有生成这样的构造函数，因此错误。

这里是关于如何处理这种情况的堆栈溢出讨论：How to force Hibernate to return dates as java.util.Date instead of Timestamp?

【讨论】：

我在 Employee 类和 BaseEmployee 中使用了 java.sql.Date。当我在 BaseEmployee 中将 java.sql.Date 更改为 java.util.Date 时，它起作用了。我刚开始学习spring web :) 我认为java.sql.date 可以在两个类中工作。谢谢你:)

【解决方案2】：

错误很简单，您的类BaseEmployee 缺少执行getBaseEmployeeById() 所需的适当构造函数。您的类使用 @NoArgsConstructor 和 @AllArgsConstructor 注释，但这些是 Spring 可能无法识别的 lombok 注释。您的方法getBaseEmployeeById() 需要一个显式定义的具有以下签名的构造函数：

public BaseEmployee(int id, Date birthDate, String firstName, String lastName, char gender, Date hireDate)

目前缺少这样的构造函数。

请在您的类中添加一个具有所述签名的构造函数，然后重试。

【讨论】：