Ошибка: javax.persistence.RollbackException: исключение [EclipseLink-4002] (Eclipse Persistence Services - 2.6.1.v20150605-31e8258)

Question

Я новичок в JPA, и у меня есть база данных UserRecords с USER_ID в качестве моего первичного ключа, и я установил для него автоинкремент.Однако всякий раз, когда я запускаю программу, она выдает мне ошибку:

javax.persistence.RollbackException: Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.6.1.v20150605-31e8258): org.eclipse.persistence.exceptions.DatabaseException
Internal Exception: java.sql.SQLSyntaxErrorException: Table/View 'USERRECORDS' does not exist.
Error Code: 30000
Call: INSERT INTO UserRecords (Email, FirstName, LastName, Password) VALUES (?, ?, ?, ?)
    bind => [4 parameters bound]
Query: InsertObjectQuery(com.Exoticatravels.UserRecords[ userId=null ])

Код для моего класса сущности:

/*
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 * and open the template in the editor.
 */
package com.Exoticatravels;

import java.io.Serializable;
import javax.persistence.Basic;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.NamedQueries;
import javax.persistence.NamedQuery;
import javax.persistence.Table;
import javax.validation.constraints.Size;
import javax.xml.bind.annotation.XmlRootElement;

/**
 *
 * @author sufiyaan
 */
@Entity
@Table(name = "UserRecords")
@XmlRootElement
@NamedQueries({
    @NamedQuery(name = "UserRecords.findAll", query = "SELECT u FROM UserRecords u")
    , @NamedQuery(name = "UserRecords.findByUserId", query = "SELECT u FROM UserRecords u WHERE u.userId = :userId")
    , @NamedQuery(name = "UserRecords.findByFirstName", query = "SELECT u FROM UserRecords u WHERE u.firstName = :firstName")
    , @NamedQuery(name = "UserRecords.findByLastName", query = "SELECT u FROM UserRecords u WHERE u.lastName = :lastName")
    , @NamedQuery(name = "UserRecords.findByEmail", query = "SELECT u FROM UserRecords u WHERE u.email = :email")
    , @NamedQuery(name = "UserRecords.findByPassword", query = "SELECT u FROM UserRecords u WHERE u.password = :password")})
public class UserRecords implements Serializable {

    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Basic(optional = false)
    @Column(name = "USER_ID")
    private Integer userId;
    @Size(max = 50)
    @Column(name = "FirstName")
    private String firstName;
    @Size(max = 50)
    @Column(name = "LastName")
    private String lastName;
    // @Pattern(regexp="[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:\\.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*@(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?", message="Invalid email")//if the field contains email address consider using this annotation to enforce field validation
    @Size(max = 100)
    @Column(name = "Email")
    private String email;
    @Size(max = 50)
    @Column(name = "Password")
    private String password;

    public UserRecords() {
    }

    public UserRecords(Integer userId) {
        this.userId = userId;
    }

    public Integer getUserId() {
        return userId;
    }

    public void setUserId(Integer userId) {
        this.userId = userId;
    }

    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;
    }

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

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

    @Override
    public boolean equals(Object object) {
        // TODO: Warning - this method won't work in the case the id fields are not set
        if (!(object instanceof UserRecords)) {
            return false;
        }
        UserRecords other = (UserRecords) object;
        if ((this.userId == null && other.userId != null) || (this.userId != null && !this.userId.equals(other.userId))) {
            return false;
        }
        return true;
    }

    @Override
    public String toString() {
        return "com.Exoticatravels.UserRecords[ userId=" + userId + " ]";
    }

}

Код для моего постоянного 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">
  <persistence-unit name="Exotica_TravelsPU" transaction-type="JTA">
    <jta-data-source>java:app/ExoticaTravels</jta-data-source>
    <exclude-unlisted-classes>false</exclude-unlisted-classes>
    <properties/>
  </persistence-unit>
</persistence>

Код для моего сервлета:

public class Register extends HttpServlet {


    @Override
    protected void doPost(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {

        String firstName = request.getParameter("firstName");
        String lastName = request.getParameter("lastName");
        String email = request.getParameter("email");
        String pass = request.getParameter("pass");

        EntityManagerFactory emf = Persistence.createEntityManagerFactory("Exotica_TravelsPU");
        EntityManager em = emf.createEntityManager();
        em.getTransaction().begin();

        UserRecords users = new UserRecords();
        users.setFirstName(firstName);
        users.setLastName(lastName);
        users.setEmail(email);
        users.setPassword(pass);

        em.persist(users);
        em.getTransaction().commit();

        em.close();
        emf.close();

        HttpSession session = request.getSession();
        session.setAttribute("firstName", firstName);

        RequestDispatcher rd = request.getRequestDispatcher("/welcome.jsp");
        rd.forward(request, response);

    }

}

Программа работает правильно, когда у меня нет USER_ID в качестве автоматического приращения.

Руководство будет очень полезно.

Заранее спасибо!