HttpGet中的android java.lang.IllegalArgumentException[重复]

Question

我正在尝试从远程服务器获取响应。下面是我的代码：

private static String baseRequestUrl = "http://www.pappico.ru/promo.php?action=";

    @SuppressWarnings("deprecation")
    public String executeRequest(String url) {
        String res = "";
        HttpClient httpClient = new DefaultHttpClient();
        HttpResponse response;      

        try {   
            //url = URLEncoder.encode(url, "UTF-8");
            HttpGet httpGet = new HttpGet(url);

            response = httpClient.execute(httpGet);
            Log.d("MAPOFRUSSIA", response.getStatusLine().toString());

            HttpEntity entity = response.getEntity();

            if (entity != null) {
                InputStream inStream = entity.getContent();
                res = streamToString(inStream);

                inStream.close();
            }
        } catch (Exception e) {
            e.printStackTrace();
        }

        return res; 
    }

    public String registerUser(int userId, String userName) {
        String res = "";
        String request = baseRequestUrl + "RegisterUser&params={\"userId\":" +
                 userId + ",\"userName\":\"" + userName + "\"}";

        res = executeRequest(request);

        return res; 
    }

在httpgethttpget=new httpGet(url)行中出现以下异常：

IllegalStateException：目标主机不能为空，或在参数中设置。scheme=null,host=null,path=http://www.pappico.ru/promo.php？action=registerUser&params={“userid”：1,“username”：“ithsoftsuve+...ithoftsuveuve....”}

不知道该怎么做才能让它发挥作用。如有任何帮助，我们将不胜感激：）

Answer 1

您必须对URL参数进行编码：

String request = baseRequestUrl + "RegisterUser&params=" +    
        java.net.URLEncoder.encode("{\"userId\":" + userId + ",\"userName\":\"" + userName + "\"}", "UTF-8");