javax.persistence.PersistenceException：名為aramis的EntityManager的持久性提供程序

Question

我是JPA和Hibernate的新手。 我正在看教程，我做了所有視頻中提到的所有事情，但我認為有些錯誤。

我正在使用eclipse java ee。 和jpa，休眠，MySQL

持久性：

<?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">
        <persistence-unit name="aramis" transaction-type="RESOURCE_LOCAL">
            <provider>org.hibernate.ejb.HibernatePersistence</provider>
            <class>com.webproject.web.entities.User</class>
            <properties>
                <property name="hibernate.connection.url" value="jdbc:mysql://localhost:3306/webproject" />
                <property name="hibernate.connection.driver_class" value="com.mysql.jdbc.Driver" />
                <property name="hibernate.connection.username" value="root" />
                <property name="hibernate.connection.password" value="mysql123" />
            </properties>
        </persistence-unit>
    </persistence>

用戶：

@Entity(name="user")
    @Table(name="user")
    public class User implements Serializable {

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

        private String name;

        @Column(nullable=false, unique=true)
                  ...

服務：公共類UserService {

    private EntityManagerFactory emf = Persistence.createEntityManagerFactory("aramis");
    private EntityManager em = emf.createEntityManager();
    private EntityTransaction tx = em.getTransaction();

    private Sha512 sha512 = new Sha512();


    public int insert(User user) {
        Date date = new Date();
        user.setCreated_at(date);
        user.setRole_type(0);

        String password = user.getPassword();
        password = sha512.encrypt(password);
        user.setPassword(password);

        tx.begin();
        em.persist(user);
        tx.commit();
        em.close();

        return user.getId();
    }

它給出了一些錯誤：

javax.persistence.PersistenceException: No Persistence provider for EntityManager named aramis
at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:61)
at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:39)
at com.webproject.web.services.UserService.<init>(UserService.java:19)
at com.webproject.web.servlet.Register.doPost(Register.java:54)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:660)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:741)

還有我的道路： 道路

Answer 1

默認情況下，當您調用Persistence.createEntityManagerFactory("aramis") ，它將嘗試在名為META-INF的文件夾中找到名為persistence.xml的文件。

為了使其工作， 您應該在類路徑中有一個META-INF/persistence.xml 。