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How to Implement One to Many Mapping in Spring Boot?

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  • Last Updated : 02 May, 2022

Spring Boot is built on the top of the spring and contains all the features of spring. Spring also provides JPA and hibernate to increase the data manipulation efficiency between the spring application and the database. In very simple terms we can say JPA (Java persistence API) is like an interface and the hibernate is the implementation of the methods of the interface Like how insertion will be down is already defined with the help of hibernate. In this article, we will discuss how to insert the values in the MySQL table using Spring JPA. Spring Initializr is a web-based tool using which we can easily generate the structure of the Spring Boot project. It also provides various different features for the projects expressed in a metadata model. This model allows us to configure the list of dependencies that are supported by JVM. Here, we will create the structure of an application using a spring initializer.

Step by Step Implementation

Step 1: Go to this link. Fill in the details as per the requirements. For this application:

Project: Maven
Language: Java
Spring Boot: 2.5.6
Packaging: JAR
Java: 11
Dependencies: Spring Web,Spring Data JPA, MySql Driver

Click on Generate which will download the starter project.

Step 2: Extract the zip file. Now open a suitable IDE and then go to File > New > Project from existing sources > Spring-boot-app and select pom.xml. Click on import changes on prompt and wait for the project to sync as pictorially depicted below as follows:

Project Structure:

Step 3: Adding the necessary properties in the file. (mapping is the database name)


Step 4: Go to src->main->java->com->example->Mapping and create two files in the Models folder i.e and Student

Project structure: table)


package com.example.Mapping.Models;
import javax.persistence.*;
// Adding the table name
@Table(name = "Address")
public class Address {
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int id;
    private String cityname;
    // Mapping the column of this table
    //Adding the name
    @JoinColumn(name = "Student_id")
    StudentInformation ob;
    Address() {}
    public Address(int id, String cityname, StudentInformation ob1)
    { = id;
        this.cityname = cityname;
        this.ob = ob1;
} by table)


// Adding the table name
@Table(name = "Student")
public class StudentInformation {
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int rollno;
    private String name;
    // Mapping to the other table
    @OneToMany(cascade = CascadeType.ALL)
    private Set<Address> ob;
    public int getRollno() { return rollno; }
    public StudentInformation() {}
    public StudentInformation(int rollno, String name)
        this.rollno = rollno; = name;
    public void setRollno(int rollno)
        this.rollno = rollno;
    public String getName() { return name; }
    public void setName(String name) { = name; }

Step 5: Adding the JPA repository of both classes in the project structure:

Project Structure:



package com.example.Mapping.Repository;
import com.example.Mapping.Models.Address;
public interface AddressRepo extends JpaRepository<Address, Integer> {



package com.example.Mapping.Repository;
import com.example.Mapping.Models.StudentInformation;
public interface StudentRepo extends JpaRepository<StudentInformation, Integer> {

Step 6: Executing the information in these tables



public class MappingApplication implements CommandLineRunner {
    @Autowired StudentRepo ob;
    @Autowired AddressRepo ob1;
    public static void main(String[] args)
    {, args);
    public void run(String... args) throws Exception
        StudentInformation student = new StudentInformation(1, "Aayush");;
        Address address = new Address(1, "Sonipat", student);;

Now run the main application

Terminal Output:


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