Send Request as JSON param inside JSON object

Question

I want to send my request parameters in JSON format. My problem is I am not able send request like below format. Can anybody help me to resolve it?

{
    "user": {
        "email" : "",
        "password": "",
        "password_confirmation": ""
    }
}

You only include example of JSON db. Please include your code that you used for sending the request as you mentioned above. — UmarZaii
– UmarZaii, Commented Jul 26, 2017 at 12:02

Naveen Tamrakar · Accepted Answer · 2017-07-26 12:11:21Z

3

Try this

      try {
            JSONObject parent = new JSONObject();
            JSONObject jsonObject = new JSONObject();

            jsonObject.put("email", "email");
            jsonObject.put("password", "password");
            jsonObject.put("password_confirmation", "password_confirmation");
            parent.put("user", jsonObject);
            Log.d("output", parent.toString());
        } catch (JSONException e) {
            e.printStackTrace();
        }

It will return this JsonObject output

{
    "user": {
        "email" : "email",
        "password": "password",
        "password_confirmation": "password_confirmation"
    }
}

edited Jul 26, 2017 at 12:11

Naveen Tamrakar

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answered Jul 26, 2017 at 12:01

King of Masses

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3 Comments

user3606971 Over a year ago

Its very useful. Thank you very much.

King of Masses Over a year ago

If it works for you, please accept the answer @user3606971

user3606971 Over a year ago

Can you help me send JSON request as below : { "gigs_list_ids" : { "2640" : 38 }, "gigs_add_ons_list_ids" : { "2641" : 65, "2642" : 66 } }

Junaid Hafeez · Accepted Answer · 2017-07-26 12:02:14Z

0

you are probably looking for something like this

try {
            JSONObject jsonObject = new JSONObject();
            jsonObject.put("email", "value");
            jsonObject.put("password", "value");
            jsonObject.put("password_confirmation", "value");

            JSONObject parent = new JSONObject();
            parent.put("user", jsonObject);
        }catch (JSONException e) {
            e.printStackTrace();
        }

answered Jul 26, 2017 at 12:02

Junaid Hafeez

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