如何将 HttpGet 转换为 HttpPost？

Question

我正在使用 HttpGet 方法从我的 Android 应用程序的 Web 服务检索数据。谁能告诉我如何将下面给出的代码转换为 HttpPost 方法？

    String url = URLEditor.encode("http://"+Constants.strURL+"Orders.asmx/CheckWebConnection?TechCode="+username+"&TechPIN="+password);
    HttpClient httpClient = new DefaultHttpClient();
    HttpGet httpGet = new HttpGet(url); 
    response = httpClient.execute(httpGet);
    HttpEntity entity = response.getEntity();
    if(entity == null) return false; 
    is = entity.getContent();

预先感谢...

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感谢您帮助我..

我尝试使用上面给出的代码。但我得到的 Document 对象为 NULL。这是我得到的 doc 为 NULL 的代码

    HttpClient httpClient = new DefaultHttpClient();
    HttpPost httpPost = new HttpPost("http://"+Constants.strURL+"Orders.asmx/CheckWebConnection");
    List<NameValuePair> nvpList = new ArrayList<NameValuePair>();
    nvpList.add(new BasicNameValuePair("TechCode", techcode));
    nvpList.add(new BasicNameValuePair("TechPIN", techpin));
    httpPost.setEntity(new UrlEncodedFormEntity(nvpList));
    HttpResponse response = httpClient.execute(httpPost);
    HttpEntity entity = response.getEntity();
    InputStream is = entity.getContent();
    DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
    DocumentBuilder db = dbf.newDocumentBuilder();
    Document doc = db.parse(is);

。我使用 HttpGet 时没有问题。怎么解决呢？请帮忙

Answer 1

这是代码，这会对您有所帮助。

ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
    nameValuePairs.add(new BasicNameValuePair("TechCode",username));
    nameValuePairs.add(new BasicNameValuePair("TechPIN",password));
try{
    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new HttpPost("http://"+Constants.strURL+"Orders.asmx/CheckWebConnection");
    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
    HttpResponse response = httpclient.execute(httppost);
    HttpEntity entity = response.getEntity();
    is = entity.getContent();
    }catch(Exception e){
    Log.e("log_tag", "Error in http connection"+e.toString());
    }