CODEWITHSUNDEEP
javajerseyjax-rs
Deepak Prabhu
Deepak Prabhu

Reputation: 219

How to add JSON object to my JSON response in Jax-rs?

My Main resource file 

public class MyResource {


@GET
@Produces(MediaType.APPLICATION_JSON)
@Path("/getAll")
public List<hotels> getResources() throws IOException {
    hotelServices services = new hotelServices();
    return services.getAllHotels();

}}

This is how I am assigning values and returning the List

public class hotelServices {

public List<hotels> getAllHotels() throws IOException

{
    List<hotels> list = new ArrayList<hotels>();    


        String ID =  "73";
        String HoteName = "KKR"
        String Location = "Kelambakkam"
        String Ratings = "2"
        String Landmark = "Opp to R&G"
        hotels thisHotel = new hotels(ID,HotelName,Location,Ratings,Landmark);
        list.add(thisHotel);
        return list;
}   }

And below is how my constructor hotels will look like

public hotels(String id, String hotelName,String location, String ratings, String landmark)
{

    this.id = id;
    this.hotelName = hotelName;
    this.location=location;
    this.ratings = ratings;
    this.landmark = landmark;
}

Am getting a response something like this

[{  
  "hotelName":" KKR",
  "id":" 73",
  "landmark":" Opp to R&G",
  "location":" Kelambakkam",
  "ratings":" 2"
}]

Am trying to generate something like this with Json objects which I am unable to do. can any help me please??

{"hotels":[{  
  "hotelName":" KKR",
  "id":" 73",
  "landmark":" Opp to R&G",
  "location":" Kelambakkam",
  "ratings":" 2"
 }]}

Upvotes: 1

Views: 623

Answers (1)

Naman
Naman

Reputation: 32046

What you're expecting is a wrapper to the List<Hotels> for which you can define a model say your already existing one - HotelServices as - 

class HotelServices {
    List<Hotels> hotels; // do all the logic of setting this value as you currently do
 }

and modify your resource as :

@GET
@Produces(MediaType.APPLICATION_JSON)
@Path("/getAll")
public HotelServices getResources() throws IOException {
    HotelServices services = new HotelServices(); // assuming you would assign value to this 
    return services;
}}

Note: Renaming by me to try and follow better naming conventions.

Upvotes: 1

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