NumberFormatException: Invalid Float "null" (Android programming, JSONObjects)

Question

For some reason I'm getting a NumberFormatException, saying "null" is not a valid float. I don't understand how that is possible, because I even test for it beforehand:

This way I put the float value into a jsonObject:

JSONObject jsonNew = new JSONObject();
    try {
        jsonNew.put("due_date", task.getDueDate().getTime());
        jsonNew.put("checked", task.isDone());
        jsonNew.put("name", task.getName());
        jsonNew.put("priority", task.getPriority());
        jsonNew.put("due_time", task.getDueTime());
        jsonNew.put("notification_before", task.getNotificationBefore());
        jsonNew.put("id", task.getId());
        jsonNew.put("proximity_alert", task.isAlertOn());
        if (task.isLocationSet()) {
            jsonNew.put("lon", task.getLongitude()+"");
            jsonNew.put("lat", task.getLatitude()+"");
        } else {
            jsonNew.put("lon", null);
            jsonNew.put("lat", null);
        }

        return jsonNew;
    } catch (JSONException e) {
        e.printStackTrace();
    }
    return null;

I am using a String here, as I learned that otherwise there might be loss when using a JSONObject. Then I try to get it afterwards:

float lat = 0, lon = 0;
            if((jsonNew.getString("lat"))!=null && (jsonNew.getString("lon"))!= null) {
                Log.d("lat = ", jsonNew.getString("lat"));
                lat = Float.parseFloat(jsonNew.getString("lat"));
                lon = Float.parseFloat(jsonNew.getString("lon"));
                locationSet = true;
            }

Log even says that it is null, but why would it even go in then? It is not my first time programming and it really bums me out that I cannot even write an easy if-statement. Can someone help out?

cygery · Accepted Answer · 2014-08-09 12:48:04Z

4

jsonNew.getString("lat") returns the String "null" and not null. Use !jsonNew.isNull("lat") etc. for the checks.

answered Aug 9, 2014 at 12:48

cygery

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1 Comment

lisa Over a year ago

Of course! Thank you! It took me so long already! It's so hard to find such easy things sometimes!

Hareshkumar Chhelana · Accepted Answer · 2014-08-09 12:48:31Z

Try to replace this peace of code,hope this will help you to solve your problem.

if((jsonNew.getString("lat"))!=null && (jsonNew.getString("lon"))!= null && !jsonNew.getString("lat").equalsIgnoreCase("null") && !jsonNew.getString("lon").equalsIgnoreCase("null")) {
    Log.d("lat = ", jsonNew.getString("lat"));
    lat = Float.parseFloat(jsonNew.getString("lat"));
    lon = Float.parseFloat(jsonNew.getString("lon"));
    locationSet = true;
}

OR

 JSONObject jsonNew = new JSONObject();
 try {
     jsonNew.put("due_date", task.getDueDate().getTime());
     jsonNew.put("checked", task.isDone());
     jsonNew.put("name", task.getName());
     jsonNew.put("priority", task.getPriority());
     jsonNew.put("due_time", task.getDueTime());
     jsonNew.put("notification_before", task.getNotificationBefore());
     jsonNew.put("id", task.getId());
     jsonNew.put("proximity_alert", task.isAlertOn());
     if (task.isLocationSet()) {
         jsonNew.put("lon", task.getLongitude()+"");
         jsonNew.put("lat", task.getLatitude()+"");
     }
    return jsonNew;
  } catch (JSONException e) {
     e.printStackTrace();
  }
  return null;


if(jsonNew.has("lat") &&  jsonNew.has("lon") && (jsonNew.getString("lat"))!=null && (jsonNew.getString("lon"))!= null) {
   Log.d("lat = ", jsonNew.getString("lat"));
   lat = Float.parseFloat(jsonNew.getString("lat"));
   lon = Float.parseFloat(jsonNew.getString("lon"));
   locationSet = true;
 }

Sush · Accepted Answer · 2014-08-09 12:58:29Z

0

Its perfectly valid that your getting the NumberFormatException. As the parseFloat java document it clearly says that if parse string is not valid then NumberFormat Exception will be thrown. In integer or float or double null is not value.

answered Aug 9, 2014 at 12:58

Sush

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