在Android中使用带有post参数的HttpClient和HttpPost

Question

我正在编写一个Android应用程序的代码,该应用程序应该获取数据,将其打包为Json并将其发布到Web服务器,而后者应该使用json进行响应.

使用GET请求工作正常,但由于某些原因使用POST,所有数据似乎都被剥离,服务器没有收到任何内容.

这是代码的片段:

HttpParams params = new BasicHttpParams();
HttpConnectionParams.setConnectionTimeout(params, 5000);
HttpConnectionParams.setSoTimeout(params, 5000);        
DefaultHttpClient httpClient = new DefaultHttpClient(params);
BasicCookieStore cookieStore = new BasicCookieStore();
httpClient.setCookieStore(cookieStore);

String uri = JSON_ADDRESS;
String result = "";
String username = "user";
String apikey = "something";
String contentType = "application/json";

JSONObject jsonObj = new JSONObject();

try {
    jsonObj.put("username", username);
    jsonObj.put("apikey", apikey);
} catch (JSONException e) {
    Log.e(TAG, "JSONException: " + e);
}

HttpPost httpPost = new HttpPost(uri);
List<NameValuePair> postParams = new ArrayList<NameValuePair>();
postParams.add(new BasicNameValuePair("json", jsonObj.toString()));
HttpGet httpGet = null;
try {
    UrlEncodedFormEntity entity = new UrlEncodedFormEntity(postParams);
    entity.setContentEncoding(HTTP.UTF_8);
    entity.setContentType("application/json");
    httpPost.setEntity(entity);

    httpPost.setHeader("Content-Type", contentType);
    httpPost.setHeader("Accept", contentType);
} catch (UnsupportedEncodingException e) {
    Log.e(TAG, "UnsupportedEncodingException: " + e);
}

try {
    HttpResponse httpResponse = httpClient.execute(httpPost);
    HttpEntity httpEntity = httpResponse.getEntity();

    if (httpEntity != null) {
        InputStream is = httpEntity.getContent();
        result = StringUtils.convertStreamToString(is);
        Log.i(TAG, "Result: " + result);
    }
} catch (ClientProtocolException e) {
    Log.e(TAG, "ClientProtocolException: " + e);
} catch (IOException e) {
    Log.e(TAG, "IOException: " + e);
}

return result;

我想我已经遵循了关于如何创建参数并发布它们的一般指导原则,但显然没有.

任何有关我可以找到解决方案的帮助或指示,都非常受欢迎(在花了几个小时意识到没有发送任何帖子数据之后).真正的服务器在Tomcat上运行Wicket,但我也在一个简单的PHP页面上测试了它,没有区别.

Answer 1

您实际上可以通过以下方式将其作为JSON发送:

// Build the JSON object to pass parameters
JSONObject jsonObj = new JSONObject();
jsonObj.put("username", username);
jsonObj.put("apikey", apikey);
// Create the POST object and add the parameters
HttpPost httpPost = new HttpPost(url);
StringEntity entity = new StringEntity(jsonObj.toString(), HTTP.UTF_8);
entity.setContentType("application/json");
httpPost.setEntity(entity);
HttpClient client = new DefaultHttpClient();
HttpResponse response = client.execute(httpPost);

Answer 2

您是否尝试过没有JSON对象并且只传递了两个basicnamevaluepairs？此外,它可能与您的服务器设置有关

更新:这是我使用的一段代码:

InputStream is = null;
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
    nameValuePairs.add(new BasicNameValuePair("lastupdate", lastupdate)); 

try {
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost(connection);
        httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
        HttpResponse response = httpclient.execute(httppost);
        HttpEntity entity = response.getEntity();
        is = entity.getContent();
        Log.d("HTTP", "HTTP: OK");
    } catch (Exception e) {
        Log.e("HTTP", "Error in http connection " + e.toString());
    }