Question

我有一个使用JPA类库的Java SE项目，问题是当它运行时有时会出现这个错误：

Exception in thread "main" java.lang.IllegalArgumentException: An exception occurred while creating a query in EntityManager: 
Exception Description: Error compiling the query [select profile from AppProfile profile where profile.user_id=?1]. Unknown entity type [AppProfile].

如果我重新运行该项目执行得很好。我的意思是有时运行正常，有时不运行

有人能说出为什么会这样吗？

我在persistence.xml中添加了这个类

com.mycompany.db.AppProfile

修改

AppUsers.class

@Entity
@Table(name="app_users")

@NamedQueries({
@NamedQuery(
    name="getAllUsers",
    query="SELECT usr FROM AppUsers usr order by usr.username asc"
),

@NamedQuery(name="findUserByUserName", 
        query="SELECT usr FROM AppUsers usr WHERE usr.username= ?1")

})

public class AppUsers implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;

    @Column(name="firstname",length=100,nullable=false)
    private String firstname;

    @Column(name="lastname",length=100,nullable=false)
    private String lastname;

    @Column(name="username",length=100,nullable=false)
    private String username;

    @Column(name="password",length=100,nullable=false)
    private String password;

    @Column(name="email",length=150,nullable=false)
    private String email;

    @Column(name="status",nullable=false)
    private int status = 0;


    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }
}

AppProfile.class

@Entity
@Table(name="app_profile")

@NamedQueries({
@NamedQuery(
    name="getUserProfile",
    query="SELECT profile FROM AppProfile profile where profile.user_id=?1"
)

})

public class AppProfile implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;


    @Column(name="user_id",nullable=false)
    private Long user_id;

    @Column(name="address",length=255,nullable=false)
    private String address;

    @Column(name="state",length=100,nullable=false)
    private String state;

    @Column(name="city",length=100,nullable=false)
    private String city;

    @Column(name="phone",length=100,nullable=false)
    private String phone;

    @Column(name="mobile",length=100,nullable=false)
    private String mobile;    

    @OneToOne(cascade=CascadeType.ALL)
    @JoinColumn(name = "user_id", insertable=false,updatable=false)
    private AppUsers userdata;



    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }


    public Long getUserId()
    {
        return this.user_id;
    }

    public AppUsers getUser()
    {
         return this.userdata;
    }

}

persistance.xml

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.0" xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd">
  <persistence-unit name="MyDBPU" transaction-type="RESOURCE_LOCAL">
    <provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>    
    <class>com.mycompany.db.AppUsers</class>
    <class>com.mycompany.db.AppProfile</class>
    <properties>
      <property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/myapp"/>
      <property name="javax.persistence.jdbc.password" value=""/>
      <property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>
      <property name="javax.persistence.jdbc.user" value=""/>
    </properties>
  </persistence-unit>
</persistence>

这是有时失败的代码

public getProfile(Long user_id)
{
    EntityManagerFactory emf=Persistence.createEntityManagerFactory("MyDBPU");
            EntityManager em=emf.createEntityManager();
            AppProfile result = null;

            try{
                EntityTransaction entr=em.getTransaction();
                entr.begin();


                String jpql="select profile from AppProfile profile where profile.user_id=?1";
                Query query=em.createQuery(jpql);
                query.setParameter(1, user_id);            


                result=(AppProfile) query.getSingleResult();
             }

            catch(NoResultException e){
                result=null;
            }

             finally{
                em.close();
             }

            return result;
}

由于

Answer 1

您在persistence.xml中包含AppProfile，但不包括AppUsers。

未知的实体类型

1 个答案: