如何将json格式数据发布到服务器

时间:2014-06-06 06:26:08

标签: android json

如何将JSON数据发布到.net服务器?

我可以使用:

            httpClient = new DefaultHttpClient();
            httpPost = new HttpPost(URL);

            List<NameValuePair> pairs = new ArrayList<NameValuePair>();
            pairs.add(new BasicNameValuePair("Code", Code));
            pairs.add(new BasicNameValuePair("Subject", Subject));
                httpPost.setEntity(new UrlEncodedFormEntity(pairs));
            httpResponse=httpClient.execute(httpPost);

或发送json数据的任何其他格式?

3 个答案:

答案 0 :(得分:2)

您可以创建JSON对象并将数据放入该对象并将其传递给服务器..

使用,

JSONObject json = new JSONObject();
json.put("UserName", "test2");
json.put("FullName", "1234567");

将数据放入json对象...

使用
json.toString();发送数据和
request.setHeader(HTTP.CONTENT_TYPE, "application/json");
标记json格式...

请参阅以下链接:

  1. Send and Receive Json android-php
  2. Sending json from Android

答案 1 :(得分:1)

以下是引用多个来源后的解决方案......

         protected Void doInBackground(Void... params) {

            httpClient = new DefaultHttpClient();
            httpPost = new HttpPost(URL);
            httpPost.setHeader("Content-Type", "application/json");

            JSONObject jsonObject = new JSONObject();

            jsonObject.put("Code", Code);
            jsonObject.put("Subject", Subject);

            stringEntity = new StringEntity(jsonObject.toString());
            stringEntity.setContentEncoding("UTF-8");
            stringEntity.setContentType("application/json");
            httpPost.setEntity(stringEntity);


            httpResponse=httpClient.execute(httpPost);  

                 return null;

           }



            protected void onPostExecute(Void result) {
              //to see the response

            httpEntity=httpResponse.getEntity();
            Response=EntityUtils.toString(httpEntity);

            System.out.println("Response "+Response);

            int Response_Code = httpResponse.getStatusLine().getStatusCode();

            System.out.println("Response_Code "+Response_Code);


            }

感谢您的回复,也帮助我找到了这个。

答案 2 :(得分:0)

您可以向名为json(或任何您想要的名称)的URL添加一个参数,并使用json字符串作为值,然后在php中获取json。

示例:

pairs.add(new BasicNameValuePair("Code", Code));
        pairs.add(new BasicNameValuePair("Subject", Subject));
pairs.add(new BasicNameValuePair("Json", my_jsons_tring));