对单个 table 策略使用鉴别器值时，第一个插入的实体的鉴别器值为空，但该值在数据库中

Question

当inheritance/single table 策略使用鉴别器值时，第一个插入实体的鉴别器值为空，但该值在数据库中。我必须重新启动服务器，以便查询结果包含鉴别器值：

package entity;

import java.io.Serializable;

import javax.persistence.Column;
import javax.persistence.DiscriminatorColumn;
import javax.persistence.DiscriminatorType;
import javax.persistence.DiscriminatorValue;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.Inheritance;
import javax.persistence.InheritanceType;
import javax.persistence.MappedSuperclass;
import javax.persistence.NamedQueries;
import javax.persistence.NamedQuery;
import javax.persistence.Table;


@Entity
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name="user_type", discriminatorType = DiscriminatorType.STRING)
@DiscriminatorValue("Null")
@Table(name="ALLUSER")
@NamedQueries({
    @NamedQuery(name = "User.findAll", query = "SELECT u FROM User u"),
    @NamedQuery(name = "User.findByAccount", query = "SELECT u FROM User u WHERE u.account = :account")
})
public class User implements Serializable{
    
    /**
     * 
     */
    private static final long serialVersionUID = 1L;
    private String account;
    private String password;
    private String userType;
    
    public User() {
        super();
    }
    public User(String account, String password) {
        super();
        this.account = account;
        this.password = password;
    }
    
    @Id
    @Column(name = "account")
    public String getAccount() {
        return account;
    }
    public void setAccount(String account) {
        this.account = account;
    }
    
    @Column(name = "password")
    public String getPassword() {
        return password;
    }
    public void setPassword(String password) {
        this.password = password;
    }
    
    @Column(name = "user_type", insertable = false, updatable = false, nullable = false)
    public String getUserType() {
        return userType;
    }
    public void setUserType(String userType) {
        this.userType = userType;
    }
    
    @Override
    public String toString() {
        return account;
    }
    
    

}
@Entity
@DiscriminatorValue("Normal")
@NamedQueries({
    @NamedQuery(name = "NormalUser.findAll", query = "SELECT u FROM NormalUser u")
})
public class NormalUser extends User implements Serializable{
    
    /**
     * 
     */
    //private String account;
    private static final long serialVersionUID = 1L;
    private LinkedHashSet<Customer> customers;
    
    public NormalUser() {
        super();
    }
    

    @OneToMany(fetch=FetchType.EAGER, mappedBy="normalUser", cascade = {CascadeType.PERSIST, CascadeType.MERGE, CascadeType.REMOVE}) //eager can be optimized when deleting a normal user
    public LinkedHashSet<Customer> getCustomers() {
        return customers;
    }

    public void setCustomers(LinkedHashSet<Customer> customers) {
        this.customers = customers;
    }
    
//  @Column(name = "account")
//  //have to override in order to get account to use
//  public String getAccount() {
//      return account;
//  }
//  
//  public void setAccount(String account) {
//      this.account = account;
//  }
}

如果我只是添加一个新的普通用户（子实体），那么查询这个用户类型为空的用户：

我使用 eclipse-link 作为 JPA 实现和 Java EE 三层 Web 架构。我知道这肯定与实体管理器的工作和持久性有关，但我不知道细节。我也不知道怎么解决。欢迎任何建议！

Answer 1

您没有在您的实体中设置 'type' 字段，并且 JPA 不会为您设置它 - 无论如何都不会在 java 对象中设置。如果在持久化实体时未设置它，则只要该实体被缓存（本地或共享 EMF 级缓存），它就会保持未设置状态。重新启动应用程序是有效的，因为它会清除缓存，强制从数据库中加载现有实体的任何提取，其中类型是根据鉴别器列值设置的。

您可以在创建class时设置类型，或者通过在实例上调用em.refresh强制从数据库重新加载数据。但是，在这种情况下，将类型列映射为基本映射似乎很奇怪 - getType 方法应该只是 return class 的静态鉴别器值，并且无论如何都不能更改类型字符串.

对单个 table 策略使用鉴别器值时，第一个插入的实体的鉴别器值为空，但该值在数据库中

When using discriminator value for single table strategy, the first inserted entity's discriminator value is null but the value is there in database

jpa

eclipselink

persistence

single-table-inheritance

discriminator