How to create correct JSONArray in Java using JSONObject

JavaJsonArrays

Java Problem Overview


how can I create a JSON Object like the following, in Java using JSONObject ?

{
	"employees": [
		{"firstName": "John", "lastName": "Doe"}, 
		{"firstName": "Anna", "lastName": "Smith"}, 
		{"firstName": "Peter", "lastName": "Jones"}
	],
	"manager": [
		{"firstName": "John", "lastName": "Doe"}, 
		{"firstName": "Anna", "lastName": "Smith"}, 
		{"firstName": "Peter", "lastName": "Jones"}
	]
}

I've found a lot of example, but not my exactly JSONArray string.

Java Solutions


Solution 1 - Java

Here is some code using java 6 to get you started:

JSONObject jo = new JSONObject();
jo.put("firstName", "John");
jo.put("lastName", "Doe");

JSONArray ja = new JSONArray();
ja.put(jo);

JSONObject mainObj = new JSONObject();
mainObj.put("employees", ja);

Edit: Since there has been a lot of confusion about put vs add here I will attempt to explain the difference. In java 6 org.json.JSONArray contains the put method and in java 7 javax.json contains the add method.

An example of this using the builder pattern in java 7 looks something like this:

JsonObject jo = Json.createObjectBuilder()
  .add("employees", Json.createArrayBuilder()
    .add(Json.createObjectBuilder()
      .add("firstName", "John")
      .add("lastName", "Doe")))
  .build();

Solution 2 - Java

I suppose you're getting this JSON from a server or a file, and you want to create a JSONArray object out of it.

String strJSON = ""; // your string goes here
JSONArray jArray = (JSONArray) new JSONTokener(strJSON).nextValue();
// once you get the array, you may check items like
JSONOBject jObject = jArray.getJSONObject(0);

Hope this helps :)

Solution 3 - Java

Small reusable method can be written for creating person json object to avoid duplicate code

JSONObject  getPerson(String firstName, String lastName){
   JSONObject person = new JSONObject();
   person .put("firstName", firstName);
   person .put("lastName", lastName);
   return person ;
} 

public JSONObject getJsonResponse(){

    JSONArray employees = new JSONArray();
    employees.put(getPerson("John","Doe"));
    employees.put(getPerson("Anna","Smith"));
    employees.put(getPerson("Peter","Jones"));

    JSONArray managers = new JSONArray();
    managers.put(getPerson("John","Doe"));
    managers.put(getPerson("Anna","Smith"));
    managers.put(getPerson("Peter","Jones"));

    JSONObject response= new JSONObject();
    response.put("employees", employees );
    response.put("manager", managers );
    return response;
  }

Solution 4 - Java

Please try this ... hope it helps

JSONObject jsonObj1=null;
JSONObject jsonObj2=null;
JSONArray array=new JSONArray();
JSONArray array2=new JSONArray();

jsonObj1=new JSONObject();
jsonObj2=new JSONObject();


array.put(new JSONObject().put("firstName", "John").put("lastName","Doe"))
.put(new JSONObject().put("firstName", "Anna").put("v", "Smith"))
.put(new JSONObject().put("firstName", "Peter").put("v", "Jones"));

array2.put(new JSONObject().put("firstName", "John").put("lastName","Doe"))
.put(new JSONObject().put("firstName", "Anna").put("v", "Smith"))
.put(new JSONObject().put("firstName", "Peter").put("v", "Jones"));

jsonObj1.put("employees", array);
jsonObj1.put("manager", array2);

Response response = null;
response = Response.status(Status.OK).entity(jsonObj1.toString()).build();
return response;

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
Questionuser2010955View Question on Stackoverflow
Solution 1 - JavaGramminView Answer on Stackoverflow
Solution 2 - JavaNaeem A. MalikView Answer on Stackoverflow
Solution 3 - JavaManasiView Answer on Stackoverflow
Solution 4 - JavaMSDView Answer on Stackoverflow