How to parse simple JSON Array in android?

Question

I've been wondering how to parse JSON Array in android ?

I've a json object like this: [{"ID":"1","email":"[email protected]","password":"password"},{"ID":"2","email":"[email protected]","password":"passward"}]

but I cannot find how to parse it, I want to access data so i could have the first mail, or the second id.

I've try around 40 differents solutions, but no luck.

Last time I tried something it was this:

private void showJSON(String json){
        try {
            JSONArray jArray = new JSONArray(json);
            String JsonString = jArray["ID"];
            tv.append(JsonString);

        } catch (JSONException e) {e.printStackTrace();}
    }

but it expected an array and not a json array. I still haven't found how to convert it into array.

I've also tried this:

for (int i = 0; i < jsonarray.length(); i++) {
    JSONObject jsonobject = jsonarray.getJSONObject(i);
    String name = jsonobject.getString("email");
}

but getJSONObject doesn't exist.

Thank you for your help.

"but getJSONObject doesn't exist" -- sure it does. Presumably, jsonarray is not a JSONArray. You may be happier using Gson, Jackson, or other more modern JSON parsing libraries. — CommonsWare
– CommonsWare, Commented Jan 5, 2016 at 22:46

Pavan Bilagi · Accepted Answer · 2016-01-06 04:04:33Z

1

Try this

private void showJSON(String json){
    try{
    JSONArray jsonArray = new JSONArray(json);
     for (int i = 0; i < jsonArray.length(); i++) {

       String id[i] = jsonArray.getJSONObject(i).getString("id");
       String email[i] = jsonArray.getJSONObject(i).getString("email");
     }
     }
    catch (JSONException e) {
                // Log.d("JSONException", e.toString());
            }
     }

answered Jan 6, 2016 at 4:04

Pavan Bilagi

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Comments

krshmbb · Accepted Answer · 2016-01-05 22:50:08Z

1

Use gson to deserialize the array back into an object. https://sites.google.com/site/gson/gson-user-guide

answered Jan 5, 2016 at 22:50

krshmbb

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Comments

Duc Nguyen · Accepted Answer · 2016-01-06 04:48:49Z

0

It is simple when using Gson: First, you create class User:

public class User
{
private String ID;
private String email;
private String password;
}

Then create UserStringJson class:

public class UserStringJson{
ArrayList<User> userArrays = new ArrayList<User>();
}

Finally, Using gson:

Gson gson = new Gson();
UserStringJson userStringJson = new UserStringJson();
userStringJson = gson.fromJson(jsonstring,
                    UserStringJson.class);
String email = userStringJson.getuserArrays().get(0).getEmail();

answered Jan 6, 2016 at 4:48

Duc Nguyen

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Comments

Rohit Suthar · Accepted Answer · 2016-01-06 04:58:13Z

Create Custom Object User :

public class User {
    public String ID;
    public String email;
    public String password;
}

Use this class into showJson() method :

private void showJSON(String json) {
    try {
        List<User> userList = new ArrayList<>();

        JSONArray jsonArray = new JSONArray(json);
        for (int i = 0; i < jsonArray.length(); i++) {
            JSONObject jsonObject = jsonArray.getJSONObject(i);
            User user = new User();
            user.ID = jsonObject.getString("id");
            user.email = jsonObject.getString("email");                 user.password = jsonObject.getString("password");
            userList.add(user);
        }
    } catch (JSONException e) {
        e.printStackTrace();
    }
}

Krishna Satwaji Khandagale · Accepted Answer · 2016-01-06 05:15:29Z

0

Do something like this:

JSONArray arr= new JSONArray("json string");
for(int i=0;i<arr.length();i++){
JSONObject obj= arr.getJSONObject(i);
// Now use this obj to get desired values.
int ID= Integer.parseInt(obj.getString("ID"))// or use getInt method
// Similarly you can do for the rest.

}

Hope it helps.

answered Jan 6, 2016 at 5:15

Krishna Satwaji Khandagale

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