Android（Java）HTTP POST JSON到服务器，但服务器告诉我没有看到POST变量

Question

我在大学的小组项目的成员为我们的应用程序创建了这个 php。

if ( $_SERVER['REQUEST_METHOD'] != 'POST' ) {
  header( 'HTTP/1.0 405 Method Not Allowed' );
  output( 2, 'only POST is accepted; received ' . $_SERVER['REQUEST_METHOD'], true );
}

# catch empty POST requests
if ( count( $_POST ) == 0 ) {
  output( 3,  'no POST variables were sent' );
}

我的代码通过了第一次测试，但第二次测试总是失败。错误代码“3”“未发送 POST 变量”

这是负责发送 POST 请求的 android 代码：

public void uploadToServer(Vector<PointOfInterest> pois)  {

        try {
            JSONObject auth = new JSONObject();
            JSONObject walk = new JSONObject();
            JSONArray points = new JSONArray();




            walk.put("walk name", "TEST NAME");
            walk.put("walk desc short", "TEST SHORT");
            walk.put("walk desc long", "TEST LONG");


            for(int i = 0; i < pois.size(); i ++){

                JSONObject location = new JSONObject();
                location.put("name", pois.get(i).getName());
                location.put("timestamp", 0);
                location.put("lattitude",pois.get(i).getLattitude());
                location.put("longitude",pois.get(i).getLongitude());
                location.put("Description",pois.get(i).getDescription());

                points.put(location);

            }

            auth.put("data", walk);

            walk.put("locations", points);

            auth.put("authorization","");
            auth.put("hash","3b6decebf0bab0e0a08c18a94849d6df1e536d65");
            auth.put("salt", "dave");

            UploadASyncTask upload = new UploadASyncTask();
            upload.execute(auth);

        } catch (Exception e) {
            e.printStackTrace();
        }
    }


private class UploadASyncTask extends AsyncTask<JSONObject, Void, Void>{

        @Override
        protected Void doInBackground(JSONObject...auth) {
            try{
                HttpClient httpclient = new DefaultHttpClient();

                HttpPost httpPost = new HttpPost("http://project.chippy.ch/upload.php");

                String json = "";

                json = auth.toString();

                StringEntity se = new StringEntity(json);
                se.setContentType(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));
                httpPost.setEntity(se);

                httpPost.setHeader("User-Agent", "WalkingTourCreator/1.0");
                httpPost.setHeader("Accept", "application/json");
                httpPost.setHeader("Content-type", "application/json");
                //httpPost.setHeader("data", json);

                HttpResponse httpResponse = httpclient.execute(httpPost);

                InputStream inputStream = httpResponse.getEntity().getContent();
                String result = "";

                if(inputStream != null){
                    result = convertInputStreamToString(inputStream);

                }
                else{
                    result = "Did not work!";

                }

                Log.d("RESULT", result);


            }catch(Exception e){

                Log.e("ERROR IN SEVER UPLOAD", e.getMessage());
            }
            return null;


        }

有什么想法，或者有什么明显的问题吗？

做php的人是这么说的：

“POST 需要键值对（据我所知）；您需要发送一个以 JSON 作为内容的“数据”键。”

据我所知，我不是这样做的吗？

我在这个网站和其他各种博客等上关注了很多帖子，他们似乎都在做与我正在做的事情相似的事情。

干杯 克里斯。

Answer 1

已解决：

HttpParams params = new BasicHttpParams();
//params.setParameter("data", auth);
HttpClient httpclient = new DefaultHttpClient(params);

HttpPost httpPost = new HttpPost("http://project.chippy.ch/upload.php");

List<NameValuePair> postParams = new ArrayList<NameValuePair>();
postParams.add(new BasicNameValuePair("data", auth.toString()));

UrlEncodedFormEntity entity = new UrlEncodedFormEntity(postParams);
entity.setContentEncoding(HTTP.UTF_8);
httpPost.setEntity(entity);

是必需的，而不是

HttpClient httpclient = new DefaultHttpClient();

HttpPost httpPost = new HttpPost("http://project.chippy.ch/upload.php");

String json = "";

json = auth.toString();

StringEntity se = new StringEntity(json);
se.setContentType(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));
httpPost.setEntity(se);

httpPost.setHeader("User-Agent", "WalkingTourCreator/1.0");
httpPost.setHeader("Accept", "application/json");
httpPost.setHeader("Content-type", "application/json");

Answer 2

你的朋友是对的，你确实需要键值对。

StringEntity 只是一个插入到 http 消息数据字段中的原始字符串。这必须正确格式化，以便接收者解析它。

尝试使用 URLEncodedFormEntity ，您将传递一个（1 个）键值对列表。键将是您要调用的参数，值将是 json 字符串

示例：

BasicNameValuePair param = new BasicNameValuePair("paramName",json);
List<BasicNameValuePair> params = new ArrayList<BasicNameValuePair>()
params.add(param);
URLEncodedFormEntity entity = new URLEncodedFormEntity(params);
entity.setContentType(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));
httpPost.setEntity(entity);

Answer 3

实际上试试这个答案。除非您像这样正确创建身份验证标头，否则尝试将身份验证作为实体字符串发送同时附加文件（即 JSON）似乎会失败：

how to send http request using http header