java - hibernate JAVA |如何在不创建新表的情况下获取EntityManager的实例？

每次我想要获取Entity Manager的实例时，都是在数据库中创建新表，那么如何在不创建新表的情况下获取Postgresql中的EntityManager（插入、删除、更新数据库）？
实体“医疗公司”：

@Entity
@Table(name = "medico")
@SequenceGenerator (name = "medico_seq",sequenceName="medico_seq",allocationSize = 1, initialValue = 1)
public class Medico implements Serializable{
    @Id
    @GeneratedValue(strategy=GenerationType.SEQUENCE,generator = "medico_seq")
    private Integer id = null;
    private String especializacao;
    private Integer diasRetorno;
    private Time tempoConsulta;
    private String nome;
    private String cpf;
    private String email;
    private String senha;
    private String telefone;
    private String cidade;
    private String estado;
    private String cep;
    private String endereco;
    private String bairro;
    private Integer enderecoNumero;


    public Medico(String especializacao, Integer diasRetorno, Time tempoConsulta, String nome, String cpf, String email, String senha, String telefone, String cidade, String estado, String cep, String endereco, String bairro, Integer enderecoNumero) {
        //...
    }
    //getters and setters
}

“麦迪科多”

package dao;

import entidade.Medico;
import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.Persistence;
import org.hibernate.Session;

public class MedicoDAO {

    public Medico salvar (Medico medico){

        EntityManager em = Persistence.createEntityManagerFactory("PersistenciaPU").createEntityManager();
        try{em.getTransaction().begin();
            if(medico.getId()==null){
                em.persist(medico); //Faz o insert
            }
            else{
                medico = em.merge(medico); // Faz o update
            }

            em.getTransaction().commit();
        }catch(Exception ex){
            System.err.println("ERRO123: "+ex.getMessage());
        }finally{
            em.close();
        }
        em.close();
        return medico;

    }
}

“持久性.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="PersistenciaPU" transaction-type="RESOURCE_LOCAL">
    <provider>org.hibernate.ejb.HibernatePersistence</provider>
    <class>entidade.HorarioAtendimentoMedico</class>
    <class>entidade.Medico</class>
    <class>entidade.Paciente</class>
    <properties>
      <property name="hibernate.cache.provider_class" value="org.hibernate.cache.NoCacheProvider"/>
      <property name="javax.persistence.jdbc.url" value="jdbc:postgresql://localhost:5432/clinica"/>
      <property name="javax.persistence.jdbc.user" value="postgres"/>
      <property name="javax.persistence.jdbc.driver" value="org.postgresql.Driver"/>
      <property name="javax.persistence.jdbc.password" value="123456"/>
      <property name="javax.persistence.schema-generation.database.action" value="create"/>
    </properties>
  </persistence-unit>
</persistence>

主要：

package persistencia;

import dao.Auxiliar;
import dao.MedicoDAO;
import entidade.Medico;
import java.sql.Time;
import javax.persistence.Persistence;

public class Persistencia {


    public static void main(String[] args) {
        Time t = new Time(2,3,4);
        Medico medico = new Medico("Cirurgiao",7,t,"Roberto","555.458.912-12","[email protected]","12345","48 3275 0463", "Florianópolis", "SC", "88.222-200","Rua Bonita","Lagoa",12);

        MedicoDAO dao = new MedicoDAO();
        dao.salvar(medico);
    }
}

错误：
线程“main”javax.persistence.persistence Exception中的异常：无法执行JPA schema generation create命令[create table medico]
由Or.PrgRESQL.UT.L.PSQLExExc:错误：关系“医学”已经存在

最佳答案

我看不到图像。但试着改变

<property name="hibernate.hbm2ddl.auto">create</property> // create new schema

对此：

<property name="hibernate.hbm2ddl.auto">update</property> // update the schema