surya surya - 9 months ago 48
Android Question

How to parse JSON GET method in android with user input parameters?

How to get data from JSON with user input parameters in android?

Answer Source

I believe the below code will give you the feasible solution for your problem,

Paste the below code in your Java file

private static JSONObject get(Context ctx, String sUrl) {
    HttpURLConnection connection = null;

    try {

        URL url = new URL(sUrl);
        connection = (HttpURLConnection) url.openConnection();
        connection.setRequestProperty("Content-Type", "application/json");
        connection.setRequestProperty("Accept", "application/json");
                "Basic " + encodedAuthentication);
        connection.setRequestProperty("Accept-Charset", "utf-8,*");
        Log.d("Get-Request", url.toString());
        try {
            BufferedReader bufferedReader = new BufferedReader(
                    new InputStreamReader(connection.getInputStream()));
            StringBuilder stringBuilder = new StringBuilder();
            String line;
            while ((line = bufferedReader.readLine()) != null) {
            Log.d("Get-Response", stringBuilder.toString());
            return new JSONObject(stringBuilder.toString());
        } finally {
    } catch (Exception e) {
        Log.e("ERROR", e.getMessage(), e);
        return null;

private static String buildSanitizedRequest(String url,
                                            Map<String, String> mapOfStrings) {

    Uri.Builder uriBuilder = new Uri.Builder();
    if (mapOfStrings != null) {
        for (Map.Entry<String, String> entry : mapOfStrings.entrySet()) {
            Log.d("buildSanitizedRequest", "key: " + entry.getKey()
                    + " value: " + entry.getValue());
    String uriString;
    try {
        uriString =; // May throw an
        // UnsupportedOperationException
    } catch (Exception e) {
        Log.e("Exception", "Exception" + e);



And your Json calling part should look like this

public static JSONObject exampleGetMethod(Context ctx, String sUrl,String yourName) throws JSONException, IOException {
    Map<String, String> request = new HashMap<String, String>();
    request.put("yourName", yourName);

    sUrl = sUrl + "yourApiName";
    return get(ctx, buildSanitizedRequest(sUrl, request));

In the above code - startDate, endDate,yourName and username are the input parameters. sUrl is the API url + API name Finally when you call exampleGetMethod(Context,String,String,String,String,String); you will get the JSON response of requested URL.

If you want to get the specific array value from the response you need to think of below logic

JSONArray a = response.getJSONArray("transactions");
JSONObject needyArray;
for (int i = 0; i < a.length(); i++) {
            needyArray = a.getJSONObject(i);

Now the needyArray JSONObject variable have the data of particular person(kumar as the example)