Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Converting a RequestBody json to an Object - Spring Boot

I am a begineer in java development but has previous experience on programming languages like PHP and Python. So little confused on how to proceed on spring boot with the development.

I am developing a rest API which has the following request

{
  "key":"value",
  "key1":"value1",
  "platform_settings":[
      {"key":"value"}
  ]
}

What I did

I created a RestController which accepts the http request and created a function for the resource

public Share share(@RequestBody final Share share) { 
        LOGGER.debug("This is the request", share);
        return share; //
}

Question 1 : If it was any other programming language like PHP or Python, there will be helper function which will accept the json request and convert it to object which I can easily work on.

In python it is as simple as

import json
import requests

response = requests.get(...)
json_data = json.loads(response.text)
//can work on json_data anyway I want.

But in java, I will have to create a POJO class, or have jackson/JPA entity as dependency which will map the request to a Class (Which I should predefine with the requests).

Is there any better way I can do this? For every request I make, I will have to create a Class which the request can be mapped to and I will have to define the class

Entity

package com.payunow.socialsharemodule.models;

import java.util.Map;

import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;


import javax.persistence.Id;


@Entity
public class Share {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;

    private String key;
    private String key1;
    private Map<String,String> platform_settings;

    public Share(String name, String description,Map<String,String> platform_settings) {
        this.key = key;
        this.key1 = key1;
        this.platform_settings = platform_settings;
    }

    //for JPA
    public Share() {}

    public String getKey() {
        return key;
    }

    public String getKey1() {
        return key1;
    }

    public Map<String,String> getPlatform_settings() {
        return platform_settings;
    }

}

For every request I make, I will have to create a class defining all its variables inside. Is this the only way to do this?

like image 317
Ajeesh Avatar asked Dec 23 '22 19:12

Ajeesh


1 Answers

You need to have Jackson dependecy for coversion of json to java object. But spring provides it by default, so you don't have to add it explicitly.

You don't need a JPA Entity. This is needed only when you want to store the recieved data into database.
Just to recieve the request you don't have to create a separate pojo class. Look at this code

@PostMapping("/json")
public JSONObject getGeneric(@RequestBody String stringToParse){
        JSONParser parser = new JSONParser();
        JSONObject json = null;
        try {
            json = (JSONObject) parser.parse(stringToParse);
        } catch (ParseException e) {
            e.printStackTrace();
        }
    return json;
}   

As you can see here it takes a string as a request and converts it into a generic JSONObject. So basically you can pass any json to this endpoint.

like image 189
pvpkiran Avatar answered Dec 26 '22 12:12

pvpkiran