Question

我试图修改下面的代码三个小时。我一直收到消息：

线程中的异常＆＃34; main＆＃34; java.lang.IllegalArgumentException：org.hibernate.hql.internal.ast.QuerySyntaxException：任务未映射[来自任务]

在JpaToDoList类中，显示警告 taskList = entityManager.createQuery（＆＃34; from task＆＃34;）。getResultList（）; 我没有，我怎么能修复项目nad有谁知道我在哪里可以找到错误？

Task.java

package test.spring.model;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.Table;


package test.spring.model;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;


@Entity
@Table( name="task")
public class Task {

@Id
@GeneratedValue(strategy=GenerationType.AUTO)
@Column(name = "id")
private int id;

@Column(name="firstname")
private String firstname;
@Column(name="lastname")
private String lastname;

    public Task() {}
    public Task(int id, String firstName, String lastName) {
        this.id = id;
        this.firstname = firstName;
        this.lastname = lastName;
    }

    public String toString() {
        return "Task#" + id + " dla " + firstname+" "+lastname;
    }

    public Task(int id) {
        this.id = id;
    }

    public int getId() {
        return id;
    }

    public void setId(int id) {
        this.id = id;
    }

    public String getFirstname() {
        return firstname;
    }

    public void setFirstname(String firstname) {
        this.firstname = firstname;
    }

    public String getLastname() {
        return lastname;
    }

    public void setLastname(String lastname) {
        this.lastname = lastname;
    }
}

的persistence.xml

<persistence version="2.0" xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd">
     <persistence-unit name="taskmanager" transaction-type="RESOURCE_LOCAL">
          <provider>org.hibernate.ejb.HibernatePersistence</provider>
          <class>test.spring.model.Task</class>          
     </persistence-unit>
</persistence>

JPA：

package test.spring.service;
import java.util.List;
import javax.persistence.EntityManager;
import javax.persistence.PersistenceContext;
import org.springframework.beans.factory.annotation.Qualifier;
import org.springframework.stereotype.Service;
import test.spring.model.Task;


@Service 
@Qualifier("jpa")
public class JpaToDoList implements ToDoListStrategy {
List<Task>taskList;

    @PersistenceContext
    private EntityManager entityManager;


    public List<Task> createListTasks() {
        taskList = (List<Task>) entityManager.createQuery("select e from Task e").getResultList();
        return taskList;
    }
}

在您修改之后，我得到下一个错误＆＃34;没有实体的默认构造函数:: test.spring.model.Task＆＃34;

以上代码包含修复程序。

任务未使用JPA映射

0 个答案: