android采用http post的方法去连接服务器的有关问题

android采用http post的方法去连接服务器的问题!
Java code
String uriPath = "http://192.168.1.82:8088/UserApp.ashx";
        InputStream result = null;
        HttpPost post = new HttpPost(uriPath);
        List<BasicNameValuePair> list = new ArrayList<BasicNameValuePair>();
        // 构建表单字段内容
        for (Entry<String, Object> entry : params.entrySet()) {
            list.add(new BasicNameValuePair(entry.getKey(), entry.getValue() + ""));
        }
        try {
            post.setEntity(new UrlEncodedFormEntity(list, HTTP.UTF_8));
            //post.addHeader("do", "Login");
            
            HttpClient client = new DefaultHttpClient();
            client.getParams().setParameter(CoreConnectionPNames.CONNECTION_TIMEOUT, CONSTANT.CONNECTION_TIMEOUT);
            client.getParams().setParameter(CoreConnectionPNames.SO_TIMEOUT, CONSTANT.SO_TIMEOUT);
            
            HttpResponse response = client.execute(post);
            //response.addHeader("Accept", "application/xml");
            
            if (CONSTANT.CODE_200 == response.getStatusLine().getStatusCode())
                result = response.getEntity().getContent();
        }
        catch (ClientProtocolException e) {
            e.printStackTrace();
        }
        catch (IOException e) {
            e.printStackTrace();
        } 
        
        if(result.equals(null))
        {
            Log.v("webserver", "多半没有取到数据");
        }
        else
            Log.v("webserver", "有数据咯");
        BufferedReader bf = new BufferedReader(new InputStreamReader(result));
        String s = null;
        try {
            s = bf.readLine();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        Log.v("webserver", s);



这个实际的uri地址应该是http://192.168.1.82:8088/UserApp.ashx?do= Login&LoginName=用户名&LoginPwd=密码
我不知道就是?do以及之后的内容怎么封装成一个地址呢?不然以我上面写的代码去请求的话就类似网页直接请求了,是非法法的参数请求,还请给位高手说下。非常感谢

------解决方案--------------------
list里面加进去不就可以了吗?
------解决方案--------------------
请求的网址是:http://192.168.1.82:8088/UserApp.ashx?do= Login&LoginName=用户名&LoginPwd=密码,想请求这个网址应该用HttpGet请求,不要用HttpPost请求。