EntityManager 没有持久性提供程序，并且数据库未更新

如何解决EntityManager 没有持久性提供程序，并且数据库未更新

我正在 Eclipse 上进行 Simple JPA 项目，以将数据插入 MysqL（版本 8.0.25）。

我遇到了这个问题，两周来我找不到解决方案，当我运行程序时显示错误：线程“main”中的异常 javax.persistence.PersistenceException: No Persistence provider for EntityManager named Eclipselink_JPA and the data不要插入到 MysqL 中的数据库中。

在 Eclipse 中，我创建了员工实体：

package com.tutorialspoint.eclipselink.entity;

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

@Entity
@Table
public class Employee {
       @Id
       @GeneratedValue(strategy = GenerationType.AUTO)  
       private int eid;
       private String ename;
       private double salary;
       private String deg;
       
       public Employee(int eid,String ename,double salary,String deg) {
              super( );
              this.eid = eid;
              this.ename = ename;
              this.salary = salary;
              this.deg = deg;
           }

           public Employee( ) {
              super();
           }

           public int getEid( ) {
              return eid;
           }
           
           public void setEid(int eid) {
              this.eid = eid;
           }
           
           public String getEname( ) {
              return ename;
           }
           
           public void setEname(String ename) {
              this.ename = ename;
           }

           public double getSalary( ) {
              return salary;
           }
           
           public void setSalary(double salary) {
              this.salary = salary;
           }

           public String getDeg( ) {
              return deg;
           }
           
           public void setDeg(String deg) {
              this.deg = deg;
           }
           
           @Override
           public String toString() {
              return "Employee [eid=" + eid + ",ename=" + ename + ",salary=" + salary + ",deg=" + deg + "]";
           }
}

我的 Persistence.xml：

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.2" 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_2.xsd">
    <persistence-unit name="Eclipselink_JPA" transaction-type="RESOURCE_LOCAL">
    <class>com.tutorialspoint.eclipselink.entity.Employee</class>
      <properties>
         <property name="javax.persistence.jdbc.url" value="jdbc:MysqL://localhost:3306/jpa_db"/>
         <property name="javax.persistence.jdbc.user" value="root"/>
         <property name="javax.persistence.jdbc.password" value="1213012"/>
         <property name="javax.persistence.jdbc.driver" value="com.MysqL.jdbc.Driver"/>
         <property name="eclipselink.logging.level" value="FINE"/>
         <property name="eclipselink.ddl-generation" value="create-tables"/>
      </properties>
    </persistence-unit>
</persistence>

创建一个名为 CreateEmloyee 的类：

package com.tutorialspoint.eclipselink.service;

import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.Persistence;
import com.tutorialspoint.eclipselink.entity.Employee;

public class CreateEmloyee {
    public static void main( String[ ] args ){
        EntityManagerFactory emfactory = Persistence.createEntityManagerFactory("Eclipselink_JPA");
          
          EntityManager entitymanager = emfactory.createEntityManager( );
          entitymanager.getTransaction( ).begin( );

          Employee employee = new Employee( ); 
          employee.setEid( 1201 );
          employee.setEname( "Gopal" );
          employee.setSalary( 40000 );
          employee.setDeg( "Technical Manager" );
          
          entitymanager.persist( employee );
          entitymanager.getTransaction( ).commit( );

          entitymanager.close( );
          emfactory.close( );
          System.out.println(employee);
    }
}

在 MysqL 数据库中，我执行了以下命令：