使用 Hibernate + MySQL 保存或更新多对多表

Question

我有 webapp，它使用 Hibernate 4、Spring 3 和 MySQL。 Hibernate 会话和事务由 spring 管理。

现在我对如何使用 Hibernate 将数据插入computer_app 表感到困惑。 下面是创建数据库的脚本：

CREATE TABLE computers (
computer_id INT AUTO_INCREMENT,
computer_name VARCHAR(15) NOT NULL,
ip_address VARCHAR(15) NOT NULL UNIQUE,
login VARCHAR(20) NOT NULL,
password VARCHAR(20) NOT NULL,
PRIMARY KEY(computer_id)
) ENGINE=InnoDB;

CREATE TABLE applications (
app_id INT AUTO_INCREMENT,
app_name VARCHAR(255) NOT NULL,
vendor_name VARCHAR(255) NOT NULL,
license_required TINYINT(1) NOT NULL,
PRIMARY KEY(app_id)
) ENGINE=InnoDB;

CREATE TABLE computer_app (
computer_id INT,
app_id INT,
FOREIGN KEY (computer_id)
    REFERENCES computers(computer_id)
    ON DELETE CASCADE,
FOREIGN KEY (app_id)
    REFERENCES applications(app_id)
    ON DELETE CASCADE
) ENGINE = InnoDB;

这里有2个对应的类，由NetBeans为computer_app表生成：

ComputerApp.java：

@Entity
@Table(name="computer_app" ,catalog="adminportal")
public class ComputerApp  implements Serializable {

    @EmbeddedId
    @AttributeOverrides( {
        @AttributeOverride(name="computerId", column=@Column(name="computer_id") ), 
        @AttributeOverride(name="appId", column=@Column(name="app_id") ) } )
     private ComputerAppId id;

    @ManyToOne(fetch=FetchType.EAGER)
    @JoinColumn(name="app_id", insertable=false, updatable=false)
     private Application applications;

    @ManyToOne(fetch=FetchType.EAGER)
    @JoinColumn(name="computer_id", insertable=false, updatable=false)
     private Computer computers;

    public ComputerApp() {
    }

    public ComputerApp(Application applications, Computer computers) {
        this.applications = applications;
        this.computers = computers;
    }

    public ComputerAppId getId() {
        return id;
    }

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

    public Application getApplications() {
        return applications;
    }

    public void setApplications(Application applications) {
        this.applications = applications;
    }

    public Computer getComputers() {
        return computers;
    }

    public void setComputers(Computer computer) {
        this.computers = computer;
    }

    @Override
    public String toString() {
        return applications.getAppName();
    }

}

ComputerAppId.java：

@Embeddable
public class ComputerAppId implements Serializable {

    @Column(name = "computer_id")
    private Integer computerId;

    @Column(name = "app_id")
    private Integer appId;

    public ComputerAppId(){

    }

    public ComputerAppId(Integer computerId, Integer appId) {
        this.computerId = computerId;
        this.appId = appId;
    }

    public Integer getComputerId() {
        return this.computerId;
    }

    public void setComputerId(Integer computerId) {
        this.computerId = computerId;
    }

    public Integer getAppId() {
        return this.appId;
    }

    public void setAppId(Integer appId) {
        this.appId = appId;
    }

    public boolean equals(Object other) {
        if ((this == other)) {
            return true;
        }
        if ((other == null)) {
            return false;
        }
        if (!(other instanceof ComputerAppId)) {
            return false;
        }
        ComputerAppId castOther = (ComputerAppId) other;

        return ((this.getComputerId() == castOther.getComputerId()) || (this.getComputerId() != null && castOther.getComputerId() != null && this.getComputerId().equals(castOther.getComputerId())))
                && ((this.getAppId() == castOther.getAppId()) || (this.getAppId() != null && castOther.getAppId() != null && this.getAppId().equals(castOther.getAppId())));
    }

    public int hashCode() {
        int result = 17;

        result = 37 * result + (getComputerId() == null ? 0 : this.getComputerId().hashCode());
        result = 37 * result + (getAppId() == null ? 0 : this.getAppId().hashCode());
        return result;
    }

}

如何使用 Hibernate 在 computer_app 数据表中的 saveOrUpdate() 数据？必须创建 2 个生成类的哪个实例 - 一个或两个？

请指出解决方案或提供一些代码..我真的需要做到这一点，直到明天！每个答案都受到高度赞赏并立即回复！如果您需要一些额外的代码 - 请告诉我。

谢谢。

Answer 1

如下定义Computer 和Application 表之间的@manytomany 关系 hibernate 将负责将记录插入到您的computer_app 表中，无需为computer_app 表定义单独的表，如下所示

电脑

@Entity
@Table(name="computers")
public class Computer {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column (name = "computer_id")
private int id;

@ManyToMany(cascade = {CascadeType.ALL},fetch=FetchType.EAGER)
@JoinTable(name="computer_app", 
        joinColumns={@JoinColumn(name="computer_id")}, 
        inverseJoinColumns={@JoinColumn(name="app_id")})
private Set<Application> applications = new HashSet<Application>();

//Setter && Getters methods

}

应用程序

@Entity
@Table(name="applications")
public class Application {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column (name = "app_id")
private int id;

@ManyToMany(mappedBy="applications",fetch=FetchType.EAGER)
private Set<Computer> computers = new HashSet<Computer>();

//Setter && Getters methods

}

保存实体

SessionFactory sf = HibernateUtil.getSessionFactory();
Session session = sf.openSession();
session.beginTransaction();

Application app1 = new Application();
Application app2 = new Application();
Computer comp = new Computer();
comp.getApplications().add(app1);
comp.getApplications().add(app2);
session.saveOrUpdate(comp);

session.getTransaction().commit();
session.close();

这将自动将记录插入所有三个表中

更多参考阅读这篇文章

Hibernate Many To Many Annotation Mapping Tutorial

希望这能解决您的问题...！

Answer 2

您只需要创建一台新计算机：

Computer c = new Computer(computer_name,ip_address,ip_address ,login, password );

和一个应用程序：

Application a = new Application(app_name,vendor_name,license_required );

然后你会这样做：

ComputerApp ca = new ComputerApp(a,c);

然后你可以像 Janus 提到的那样坚持它。 Hibernate 将处理外键，因为您将在构造函数中将计算机 c 和应用程序 a 作为参数传递

Answer 3

您可以使用 EntityManager 保存对象：

public void save(ComputerApp t){
// begin transaction 
em.getTransaction().begin();
if (!em.contains(t)) {
    // persist object - add to entity manager
    em.persist(t);
    // flush em - save to DB
    em.flush();
}
// commit transaction at all
em.getTransaction().commit();

}