Question

我对hibernate 4.x中的一对多映射存在问题，我使用的是mysql5.6。

首先让我们看看我的2个实体，第一个用户实体，

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;
    }





}

第二实体-Booking Class

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;
    }

现在问题是一个用户可以进行多次预订。因此代码在预订表中创建外键，该预订表是预订的用户密钥。但是当一个用户预订多个预订时，会给出唯一约束错误。所以我制作了不符合我需要的用户自动生成器，因为如果同一个用户预订两个预订，那么userid应该是相同的，但是当我这样做时，它给了我唯一的约束错误。

请告诉我该如何实施。

由于

Answer 1

第一件事 - 用户的ID不应该是自动生成的，也许肯定是唯一的移动号码应该是标识符

第二件事 - 当您添加预订时，您将检查用户是否存在，是否存在将预订添加到用户，否则创建用户并添加预订，然后提交用户。

试试这些并告诉我。

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
       }
   }

Hibernate关系一对多映射问题

1 个答案: