JSON Android跳过尝试并获取捕获消息

Question

您好我正在尝试为我的应用编写一个log_in按钮，这样它在另一个应用上工作正常，但它无法正常工作，它向我显示JSONException捕获中的消息

这是我的代码

try {

    JSONObject json=null;
    JSONArray jArray = new JSONArray(result);
    for( int i=0;i<jArray.length();i++){
        json = jArray.getJSONObject(i); 
        Toast.makeText(getApplicationContext(),"Work fine man " + json.getString("name"),Toast.LENGTH_LONG).show();
       Intent s = new Intent(Sign_in.this, Search.class);
        startActivity(s);

    }
}
catch(JSONException e)
{
    Toast.makeText(getApplicationContext(), "Wrong Mail and pass",Toast.LENGTH_LONG).show();    
}

这是result

String result ="";
try{
    BufferedReader reader = new BufferedReader(new InputStreamReader(is));  
    StringBuilder sb = new StringBuilder();
    String line = null;
    while((line = reader.readLine()) !=null){
        sb.append(line +"\n");
    }
    is.close();
    result=sb.toString();
}
catch(Exception e){
    Log.e("log_tag", "erorr converting result "+e.toString());  
}