1

我是 android 的新手，并试图通过发送 json 对象来使用 .net weservice。

我正在创建一个 json 对象并向其中添加所有参数，例如

 JSONObject json = new JSONObject();

   json.put("name","koli"); 
   json.put("email","asdf@email.com");

并在 AsyncTask 类的 doInBackground 方法中调用请求

String response = HttpClient.SendHttpPost(params[0],json);

这就是我试图提出请求的方式

public static String SendHttpPost(String URL, JSONObject jsonData) {

        Log.d("jsonData", ""+jsonData);
        StringBuilder stringBuilder = new StringBuilder();
        DefaultHttpClient client = new DefaultHttpClient();
        HttpPost httpPostRequest = new HttpPost(URL);
        httpPostRequest.setHeader("User-Agent", "com.altaver.android_PostJson2");
        httpPostRequest.setHeader("Accept", "application/json");
        httpPostRequest.setHeader("Content-Type", "application/json");
       StringEntity se = null;
        try {
            se = new StringEntity(jsonData.toString(),"UTF-8");

            se.setContentType("application/json");
        } catch (UnsupportedEncodingException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        httpPostRequest.setEntity(se);
        HttpResponse response = null;
        try {
            response = client.execute(httpPostRequest);
        } catch (ClientProtocolException e) {
            // TODO Auto-generated catch block

            e.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

在上述方法中

StringEntity se = null;
            try {
                se = new StringEntity(jsonData.toString(),"UTF-8");

                se.setContentType("application/json");
            } catch (UnsupportedEncodingException e) {

这个片段会做。

为什么我无法发送我的 jsonobject？

谢谢：）

4

2 回答 2

2

试试这个..在你的 doInBackground

JSONObject jObjOut = null;

                try {
                    HttpPost request = new HttpPost(url);
                    JSONObject returnedJObject;
                    returnedJObject = new JSONObject(JSONdata.toString());
                    JSONStringer json = new JSONStringer();
                    StringBuilder sb=new StringBuilder();


                    if (returnedJObject!=null)
                    {
                        Iterator<String> itKeys = returnedJObject.keys();
                        if(itKeys.hasNext())
                            json.object();
                        while (itKeys.hasNext())
                        {
                            String k=itKeys.next();
                            json.key(k).value(returnedJObject.get(k));
                            //Log.e("keys "+k,"value "+returnedJObject.get(k).toString());
                        }             
                    }
                   json.endObject();


                   StringEntity entity;

                   entity = new StringEntity(json.toString());

                   entity.setContentType("application/json;charset=UTF-8");
                  // entity.setContentEncoding(new BasicHeader(HTTP.CONTENT_TYPE,"application/json;charset=UTF-8"));
                   request.setHeader("Accept", "application/json");
                   request.setEntity(entity);

                   HttpResponse response =null;
                   DefaultHttpClient httpClient = new DefaultHttpClient();

                   HttpConnectionParams.setSoTimeout(httpClient.getParams(), 30000);
                   HttpConnectionParams.setConnectionTimeout(httpClient.getParams(),50000);

                   response = httpClient.execute(request);

                   InputStream in;
                   in = response.getEntity().getContent();

                   BufferedReader reader = new BufferedReader(new InputStreamReader(in));
                   String line = null;
                   while((line = reader.readLine()) != null){
                       sb.append(line);
                   }


                   return new JSONObject(sb.toString());
于 2013-10-08T08:50:45.783 回答
1

尝试更换：

se = new StringEntity(jsonData.toString(),"UTF-8");

对此：

se = new StringEntity(jsonData.toString());
于 2013-10-08T09:09:12.013 回答