无法在JPA控制台Intellij中执行JPQL查询
我是JavaEE新手,很难在Intellij中使用JPA控制台。在执行JPQL查询时浪费了大量时间来消除此错误:无法在JPA控制台Intellij中执行JPQL查询,jpa,intellij-idea,jpql,Jpa,Intellij Idea,Jpql,我是JavaEE新手,很难在Intellij中使用JPA控制台。在执行JPQL查询时浪费了大量时间来消除此错误: jpa-ql> SELECT p FROM PersonEntity p [2017-04-21 18:44:46] using C:\Users\admin\.IntelliJIdea2017.1\system\compiler\jpaproject4.0.82b748b3\.generated\Jpa_Console\JpaProject4.0-PU-14927894866
jpa-ql> SELECT p FROM PersonEntity p
[2017-04-21 18:44:46] using C:\Users\admin\.IntelliJIdea2017.1\system\compiler\jpaproject4.0.82b748b3\.generated\Jpa_Console\JpaProject4.0-PU-1492789486636\META-INF\persistence.xml
[2017-04-21 18:44:48] PersonEntity is not mapped [SELECT p FROM PersonEntity p]
我在MySQL中创建了简单的JavaEE持久化项目和简单数据库。
我在Intellij中添加了数据源并生成了持久性映射。
我导入了必要的库并消除了所有编译错误。
我可以在数据库工具窗口中看到我的数据库,并可以执行SQL查询。
但由于这个恼人的错误,我无法执行JPQL查询:PersonEntity没有映射
persistence.xml:
<persistence-unit name="NewPersistenceUnit">
<provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
<class>models.PersonEntity</class>
<properties>
<property name="hibernate.connection.url" value="jdbc:mysql://localhost:3306/sample"/>
<property name="hibernate.connection.driver_class" value="com.mysql.jdbc.Driver"/>
<property name="hibernate.connection.username" value=""/>
<property name="hibernate.connection.password" value=""/>
<property name="hibernate.archive.autodetection" value="class"/>
<property name="hibernate.show_sql" value="true"/>
<property name="hibernate.format_sql" value="true"/>
<property name="hbm2ddl.auto" value="update"/>
</properties>
</persistence-unit>
数据库结构:
项目结构:
有什么办法解决这个问题吗?我自己设法解决了这个问题
授予*上的所有权限*到“根”@“本地主机”
import models.personetity;
导入org.hibernate.jpa.HibernatePersistenceProvider;
导入javax.persistence.EntityManager;
导入javax.persistence.EntityManagerFactory;
导入javax.persistence.spi.PersistenceProvider;
导入java.util.HashMap;
导入java.util.List;
公共类应用程序{
public static final String QUERY=“从PersonEntity p中选择p”;
公共静态void main(字符串[]args){
PersistenceProvider PersistenceProvider=新的HibernatePersistenceProvider();
EntityManagerFactory=persistenceProvider.createEntityManagerFactory(“NewPersistenceUnit”,newHashMap());
EntityManager em=factory.createEntityManager();
List resultList=em.createQuery(QUERY,PersonEntity.class).getResultList();
System.out.println(结果列表);
em.close();
}
}`
package models;
import javax.persistence.*;
@Entity
@Table(name = "person", schema = "sample", catalog = "")
public class PersonEntity {
private int id;
private String name;
@Id
@Column(name = "id")
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
@Basic
@Column(name = "name")
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
PersonEntity that = (PersonEntity) o;
if (id != that.id) return false;
if (name != null ? !name.equals(that.name) : that.name != null) return false;
return true;
}
@Override
public int hashCode() {
int result = id;
result = 31 * result + (name != null ? name.hashCode() : 0);
return result;
}
}
import models.PersonEntity;
import org.hibernate.jpa.HibernatePersistenceProvider;
import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.spi.PersistenceProvider;
import java.util.HashMap;
import java.util.List;
public class App {
public static final String QUERY = "SELECT p FROM PersonEntity p";
public static void main(String[] args) {
PersistenceProvider persistenceProvider = new HibernatePersistenceProvider();
EntityManagerFactory factory = persistenceProvider.createEntityManagerFactory("NewPersistenceUnit", new HashMap());
EntityManager em = factory.createEntityManager();
List<PersonEntity> resultList = em.createQuery(QUERY, PersonEntity.class).getResultList();
System.out.println(resultList);
em.close();
}
}`