java - 为什么我的 EntityManager Helper 类返回以前的查询结果以及如何修复它？

Question

这个问题What is the right way to use entitymanager 提供了 EntityManagerHelper.java 的链接我将其添加到我的代码库中。当我使用此帮助程序类对数据库进行后续调用时，它会将先前的结果返回到同一查询。

我最常看到的场景是检索我的用户类的lastentry属性。在浏览器上，我发出 AJAX 请求，下面的 servlet 部分获取用户对象并调用一个方法来返回我的 lastentry。

我已经阅读过有关 .clear() 的内容，但当我将其添加到 EntityManagerHelper 时，出现服务器错误。我想避免每次想要调用数据库时都创建 EntityManager。

如何解决这个问题？

部分来自 servlet

User user = User.getUser();

response.setContentType("application/json");
response.setCharacterEncoding("UTF-8");
response.getWriter().write(user.getLastentry());

用户类别

package entities;

import java.io.Serializable;
import java.util.Date;
import javax.persistence.Basic;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.NamedQueries;
import javax.persistence.NamedQuery;
import javax.persistence.Persistence;
import javax.persistence.Table;
import javax.persistence.Temporal;
import javax.persistence.TemporalType;
import javax.validation.constraints.NotNull;
import javax.validation.constraints.Size;
import javax.xml.bind.annotation.XmlRootElement;
import org.apache.shiro.SecurityUtils;
import responseablees.EntityManagerHelper;

/**
 *
 * @author Christopher Loughnane <chrisloughnane1@gmail.com>
 */
@Entity
@Table(name = "user")
@XmlRootElement
@NamedQueries({
    @NamedQuery(name = "User.findAll", query = "SELECT u FROM User u")
    , @NamedQuery(name = "User.findById", query = "SELECT u FROM User u WHERE u.id = :id")
    , @NamedQuery(name = "User.findByUsername", query = "SELECT u FROM User u WHERE u.username = :username")
    , @NamedQuery(name = "User.getUsernameByUserEmail", query = "SELECT u.username FROM User u WHERE u.useremail = :useremail")
    , @NamedQuery(name = "User.findByUserEmail", query = "SELECT u FROM User u WHERE u.useremail = :useremail")
    , @NamedQuery(name = "User.findByPassword", query = "SELECT u FROM User u WHERE u.password = :password")})
public class User implements Serializable {

    @Basic(optional = false)
    @NotNull
    @Size(min = 1, max = 2048)
    @Column(name = "lastentry")
    private String lastentry;

    @Basic(optional = false)
    @NotNull
    @Size(min = 1, max = 128)
    @Column(name = "useremail")
    private String useremail;
    @Basic(optional = false)
    @Column(name="created", insertable = false, updatable = false)
    @Temporal(TemporalType.TIMESTAMP)
    private Date created;

    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Basic(optional = false)
    @Column(name = "id")
    private Integer id;
    @Basic(optional = false)
    @NotNull
    @Size(min = 1, max = 100)
    @Column(name = "username")
    private String username;
    @Basic(optional = false)
    @NotNull
    @Size(min = 1, max = 100)
    @Column(name = "password")
    private String password;

    public User() {
    }

    public User(Integer id) {
        this.id = id;
    }

    public User(Integer id, String username, String password) {
        this.id = id;
        this.username = username;
        this.password = password;
    }

    public Integer getId() {
        return id;
    }

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

    public String getUsername() {
        return username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    public String getPassword() {
        return password;
    }

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

    @Override
    public int hashCode() {
        int hash = 0;
        hash += (id != null ? id.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 User)) {
            return false;
        }
        User other = (User) object;
        if ((this.id == null && other.id != null) || (this.id != null && !this.id.equals(other.id))) {
            return false;
        }
        return true;
    }

    @Override
    public String toString() {
        return "DAOs.User[ id=" + id + " ]";
    }

    public String getUseremail() {
        return useremail;
    }

    public void setUseremail(String useremail) {
        this.useremail = useremail;
    }

    public Date getCreated() {
        return created;
    }

    public void setCreated(Date created) {
        this.created = created;
    }

    public String getLastentry() {
        return lastentry;
    }

    public void setLastentry(String lastentry) {
        this.lastentry = lastentry;
    }

    public static User getUser(){

        String currentUser = (String) SecurityUtils.getSubject().getPrincipal();
        User user = (User) em.createNamedQuery("User.findByUserEmail")
                .setParameter("useremail", currentUser)
                .getSingleResult();
        return user;
    }
}

请求的代码

详细方法

    String currentUser = (String) SecurityUtils.getSubject().getPrincipal();

    EntityManagerFactory emfactory = Persistence.createEntityManagerFactory("com.mycompany_responseableES_war_1.0-SNAPSHOTPU");
    EntityManager em = emfactory.createEntityManager();

    User user = (User) em.createNamedQuery("User.findByUserEmail")
            .setParameter("useremail", currentUser)
            .getSingleResult();
    em.getTransaction().begin();
    user.setLastentry(JSON);
    em.getTransaction().commit();

EntityManagerHelper方法

    User user = User.getUser();
    EntityManager em = EntityManagerHelper.getEntityManager();
    em.getTransaction().begin();
    user.setLastentry(JSON);
    em.getTransaction().commit();

Answer 1

首先，我会小心地将所有我的数据获取/修改代码放在

em.getTransaction().begin()

和

em.getTransaction().commit()

这可以确保您的实体受到管理，并且您的更改自动保留。在您的代码中，User 实例看起来来自事务外部。我没有太多使用 JPA 和手动事务，但如果用户分离(即不由 EntityManager 管理)，我不会感到惊讶，因此更改是不会自动持久化到数据库。如果是这种情况，我想你可以通过放置来解决它

// merge(...) takes an entity that is not managed, apply changes to
// the persistence context and returns a new *managed* entity.
// The original entity REMAINS *detached* or *non persistent*
User managedUser = em.merge(user);

// I assume that you have something like this to set your user in the
// current session or whatever
User.setUser(managedUser);

就在提交之前。

JPA 词汇一开始可能有点困难，请阅读本文以了解有关 the JPA entity lifecycle 的更多详细信息。 (特别是底部的图表)。