Question

错误就像项目启动时一样 “org.apache.jasper.JasperException：java.lang.IllegalArgumentException：Object：emp.Customer [id = 100]不是已知的实体类型。” 如何解决这个错误请告诉我

1）newjsp.jsp

<%-- 
    Document   : newjsp
    Created on : Oct 10, 2015, 1:10:21 PM
    Author     : Niral
--%>`enter code here`

<%@page import="emp.*"%>
<%@page import="javax.persistence.Persistence"%>
<%@page import="javax.persistence.EntityManager"%>
<%@page import="javax.persistence.EntityManagerFactory"%>
<%@page contentType="text/html" pageEncoding="UTF-8"%>
<!DOCTYPE html>
<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <title>JSP Page</title>
    </head>
    <body>
        <%
             EntityManagerFactory emf = Persistence.createEntityManagerFactory("WebApplicationJPAPU");
                EntityManager em = emf.createEntityManager();
                //javax.transaction.UserTransaction utx;

                Customer c1 = new Customer();
                em.getTransaction().begin();
                c1.setId(100l);
                c1.setBalance(10000);

                em.persist(c1);
                em.getTransaction().commit();
        %>
    </body>
</html>

2)

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

package emp;

import java.io.Serializable;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.*;
import javax.persistence.Id;

/**
 *
 * @author Niral
 */
@Entity
@Table(name = "CustMaster")
public class Customer implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "CustId")
    private Long id;
    @Column(name = "CustBalance")
    private double Balance;

    public double getBalance() {
        return Balance;
    }

    public void setBalance(double Balance) {
        this.Balance = Balance;
    }

    public Long getId() {
        return id;
    }

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

    @Override
    public int hashCode() {
        int hash = 0;
        hash += (id != null ? id.hashCode() : 0);
        return hash;
    }



    @Override
    public String toString() {
         return "emp.Customer[ id=" + id + " ]";
    }


}

3)persistence.xml


<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.1" xmlns="http://xmlns.jcp.org/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd">

<provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>   
    <persistence-unit name="WebApplicationJPAPU" transaction-type="JTA">
    <jta-data-source>jdbc/e4</jta-data-source>
    <exclude-unlisted-classes>false</exclude-unlisted-classes>
    <properties>
      <property name="javax.persistence.schema-generation.database.action" value="drop-and-create"/>
    </properties>`enter code here`
  </persistence-unit>
</persistence>

Answer 1

您的问题可能类似于post。

理想情况下，由于您在persistence.xml文件中注释了<exclude-unlisted-classes>false</exclude-unlisted-classes>，因此它应该自动检测类路径中的实体类。问题是，出于某种原因它没有这样做。

如果您只有一小组实体类，或假设您只有一个实体需要管理（在本例中是您的客户实体），那么我建议您明确列出persistence.xml文件中的实体类。见下面的例子：

<persistence-unit name="WebApplicationJPAPU" transaction-type="JTA">
    <class>emp.Customer</class>
    ...
</persistence-unit>

org.apache.jasper.JasperException：java.lang.IllegalArgumentException：Object：emp.Customer [id = 100]不是已知的实体类型

1 个答案: