休眠關系一對多映射問題

Question

我在hibernate 4.x中有一個關於一對多映射的問題，我正在使用mysql5.6。

首先，請向您展示我的2個實體，第一個用戶實體，

package com.project.entities;

import java.util.ArrayList;
import java.util.Collection;

import javax.persistence.CascadeType;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.OneToMany;
import javax.persistence.Table;

@Entity
@Table(name="UserBookingEntryTable")
public class UserBookingEntryClass {

    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    private int uid;
    private String ubmobile_no;
    private String ubname;
    @OneToMany(cascade = {CascadeType.ALL},mappedBy="user")
    private Collection<BookingServicesClass> bookings=new ArrayList<BookingServicesClass>();


    public Collection<BookingServicesClass> getBookings() {
        return bookings;
    }
    public void setBookings(Collection<BookingServicesClass> bookings) {
        this.bookings = bookings;
    }
    public int getUid() {
        return uid;
    }
    public void setUid(int uid) {
        this.uid = uid;
    }




    public String getUbmobile_no() {
        return ubmobile_no;
    }
    public void setUbmobile_no(String ubmobile_no) {
        this.ubmobile_no = ubmobile_no;
    }
    public String getUbname() {
        return ubname;
    }
    public void setUbname(String ubname) {
        this.ubname = ubname;
    }





}

第二實體預訂類

package com.project.entities;

import java.util.Date;

import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;
import javax.persistence.Table;

@Entity
@Table(name = "BookingServiceTable")
public class BookingServicesClass {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private int booking_id;

    private String first_name;

    private String last_name;

    private String mobile;

    private String location;

    private String booking_address;

    private String booking_type;

    private String landmark;
    private int booking_pincode;
    private Date booking_date;
    @ManyToOne
    @JoinColumn(name="usid")
    private UserBookingEntryClass user;
    public Integer getBooking_id() {
        return booking_id;
    }

    public void setBooking_id(Integer booking_id) {
        this.booking_id = booking_id;
    }

    public String getFirst_name() {
        return first_name;
    }

    public void setFirst_name(String first_name) {
        this.first_name = first_name;
    }

    public String getLast_name() {
        return last_name;
    }

    public void setLast_name(String last_name) {
        this.last_name = last_name;
    }

    public String getMobile() {
        return mobile;
    }

    public void setMobile(String mobile) {
        this.mobile = mobile;
    }

    public String getLocation() {
        return location;
    }

    public void setLocation(String location) {
        this.location = location;
    }

    public String getBooking_address() {
        return booking_address;
    }

    public void setBooking_address(String booking_address) {
        this.booking_address = booking_address;
    }

    public String getLandmark() {
        return landmark;
    }

    public void setLandmark(String landmark) {
        this.landmark = landmark;
    }

    public Integer getBooking_pincode() {
        return booking_pincode;
    }

    public void setBooking_pincode(Integer booking_pincode) {
        this.booking_pincode = booking_pincode;
    }

    public Date getBooking_date() {
        return booking_date;
    }

    public void setBooking_date(Date booking_date) {
        this.booking_date = booking_date;
    }

    public String getBooking_type() {
        return booking_type;
    }

    public void setBooking_type(String booking_type) {
        this.booking_type = booking_type;
    }

    public UserBookingEntryClass getUser() {
        return user;
    }

    public void setUser(UserBookingEntryClass user) {
        this.user = user;
    }


}

    // to add booking from a user:
    public String addService(String fname, String lname, String mob,
            String ser, String loc, String add, String lm, int pc, String bd) {
        String res = "failure";
        SimpleDateFormat sdf = new SimpleDateFormat("MM/dd/yyyy");
        Date convertedCurrentDate = null;

        try {
            convertedCurrentDate = sdf.parse(bd);
        } catch (ParseException e1) {
            // TODO Auto-generated catch block
            e1.printStackTrace();
        }
        BookingServicesClass bs = new BookingServicesClass();
        UserBookingEntryClass ubs = new UserBookingEntryClass();
        Session session = HibernateUtil.getSessionFactory().openSession();

        session.beginTransaction();
        System.out.println("dddd");
        try {


            **//userid is auto generated**
            ubs.setUbmobile_no(mob);
            ubs.setUbname((fname + " " + lname));
            // bs.setBid(1);
            bs.setFirst_name(fname);
            bs.setLast_name(lname);
            bs.setMobile(mob);
            bs.setLocation(loc);
            bs.setBooking_type(ser);
            bs.setBooking_address(add);
            bs.setLandmark(lm);
            bs.setBooking_pincode(pc);
            bs.setBooking_date(convertedCurrentDate);
            bs.setUser(ubs);
            session.save(ubs);
            session.save(bs);

            // Commit the transaction
            session.getTransaction().commit();
            /*session.close();*/
            res = "success";
            System.out.println("ajax");

        } catch (Exception e) {
            System.out.println(e);
            session.getTransaction().rollback();
            res = "failure";
            // success=false;
        }

        return res;
    }

---

現在的問題是，一個用戶可以進行多個預訂。 因此，代碼在預訂表中創建了一個外鍵，即要預訂的用戶鍵。 但是，當一個用戶預訂多個預訂時，將給出唯一約束錯誤。 所以我使userid自動生成不符合我的需求，因為如果同一用戶預訂兩次預訂，則userid應該相同，但是當我這樣做時，它給了我獨特的約束錯誤。

請告訴我該如何實施。

謝謝

Answer 1

第一件事-不應自動生成User的ID，也許絕對唯一的mobilenumber應該是標識符

第二件事-添加預訂時，您將檢查用戶是否存在，如果存在，則向用戶添加預訂，否則創建用戶並添加預訂，然后提交用戶。

試試這些，讓我知道。

BookingServicesClass  booking = new BookingServicesClass ()
if(ubmobile_no!=null) {
    UserBookingEntryClass user= session.get(UserBookingEntryClass .class, ubmobile_no);
    if (user!= null) {
          booking.setUser(user);
          user.getBookings().add(booking);
       } else {
          //do nothing
       }
   }