How to create JAVA object from JSON string?

Question

I am creating an app that makes a GET Request to an API and receives a JSON and I need to create class that holds the information. This is what I have tried :

package apiwebapprequest;

import com.google.gson.Gson;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.URL;
import java.net.URLConnection;


public class APIWebAppRequest {


public static void main(String[] args) throws IOException {
    Gson gson = new Gson();
    Object obj = new Object();

    try {   
 URL url = new URL("xxx");
 URLConnection yc = url.openConnection();
 BufferedReader in = new BufferedReader(new InputStreamReader(yc.getInputStream()));
 String inputLine;

 while((inputLine = in.readLine())!= null){
     System.out.println(inputLine);

     in.close();
     gson.toJson(obj,inputLine);




 }

 }catch(Exception e) {System.out.println(e);}



}

}

This is the object class:

package apiwebapprequest;

public class Object {


private String cif;
private String data_creare;

public String getCif() {
    return cif;
}

public void setCif(String cif) {
    this.cif = cif;
}

public String getData_creare() {
    return data_creare;
}

public void setData_creare(String data_creare) {
    this.data_creare = data_creare;
}




}

On line 29 - > gson.toJson(obj,inputLine); it gives me an error. Can you please tell me how to do it ? I can`t find how to fix it or modify in other way that it works

The ideea is that i get out when i make request in inputLine I have json and I want to save the fields separate in Object properties

"it gives me an error" - are we supposed to guess the error? It would be easier to help you if you'd provide the error message for us. — Amongalen
– Amongalen, Commented Jan 27, 2020 at 10:22
Other than that, there is a lot of resources online on how to read/parse JSON in Java, eg. this article — Amongalen
– Amongalen, Commented Jan 27, 2020 at 10:25

Ganesh Gudghe · Accepted Answer · 2020-01-27 10:33:35Z

1

You can use the following way.

gson.toJson(obj); // is used to convert object to JSON

if you want to convert JSON to Java Object then you can use

gson.fromJson(json, Car.class);

Ex.

     public class Car {
            public String brand = null;
            public int    doors = 0;
   // add getter and setter

        }

String json = "{\"brand\":\"Jeep\", \"doors\": 3}";

Gson gson = new Gson();

Car car = gson.fromJson(json, Car.class);

answered Jan 27, 2020 at 10:33

Ganesh Gudghe

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2 Comments

SECI Over a year ago

you hard-coded the values, how can i take from variable that contains json ?

Rörd Over a year ago

@SECI If you don't know the structure of the JSON data beforehand, you can use the low-level Parser API of Gson, e.g. JsonParser.parseString(jsonString) will return a JsonElement object which you can then analyze further.

Smile · Accepted Answer · 2020-01-27 10:23:00Z

1

You should read the entire response first in the while loop and then convert to json after while loop ends. You can modify your code something like

StringBuilder sb = new StringBuilder();
while((inputLine = in.readLine())!= null){
     System.out.println(inputLine);
     sb.append(inputLine +"\n");
     in.close();
 }
gson.toJson(obj,sb.toString());

answered Jan 27, 2020 at 10:23

Smile

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2 Comments

SECI Over a year ago

It says no suitable method found for toJson , do you know how to use Gson can you please help me with the correct method ? Thanks in advance

Smile Over a year ago

You can do SoModel model = gson.fromJson(sb.toString(), SoModel.class); where SoModel is the java model class similar to structure of json response of service

Sukreet Roy Choudhury · Accepted Answer · 2020-01-27 10:26:59Z

0

Try using gson.toJson(obj,inputLine) after the while loop.

Suppose the json is

{
"foo":"bar",
"foo1":"bar1"
}

Each line is not a valid JSON. Hence you need to read the entire string 1st and then parse it.

answered Jan 27, 2020 at 10:26

Sukreet Roy Choudhury

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Comments

Rakesh · Accepted Answer · 2020-01-27 10:32:23Z

0

You can use Jackson class com.fasterxml.jackson.databind.ObjectMapper.

Following is how you can parse the json to the class:

ObjectMapper objectMapper = new ObjectMapper();
MyClass myClass = objectMapper.readValue(json, MyClass.class);

answered Jan 27, 2020 at 10:32

Rakesh

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