Question

从android我希望将学生详细信息作为json String发送到服务器。所以，我有： -

String url1 = "http://my ipaddress/AndroidProject/details";

List<Pair<String, String>> pairs = new ArrayList<>();
pairs.add(new Pair<>("details", JsonString));

URL url = new URL(url1);
HttpsURLConnection conn = (HttpsURLConnection) url.openConnection();
conn.setRequestMethod("POST"); 
conn.setDoInput(true); 
conn.setDoOutput(true); 

// POST the signup
try {
OutputStream os = conn.getOutputStream();
BufferedWriter writer = new BufferedWriter( new OutputStreamWriter(os, "UTF-8"));       
writer.write(pairs.toString());
writer.flush(); 
writer.close(); 
os.close(); 
conn.connect(); 
} 
catch (IOException ioe) {

//error display

}

而且，接收我有

@RequestMapping(value = "/details", method = RequestMethod.POST, produces = "application/json")
    public ResponseEntity<Object> processUserdetails(HttpServletRequest request, HttpServletResponse response,
            @RequestParam(value = "details", required = true) final String studentDetail) {

        logger.info("Mobile Student Signup > Message received: " + studentDetail);
}

但是服务器没有收到任何数据..为什么？

Answer 1

尝试使用这种方式：

//Activity Class variable
StringBuilder sb;   //don't forget to initialize it!

private class RegisterDevice extends AsyncTask<String,String,String>{
    @Override
    protected void onPreExecute() {
        super.onPreExecute();
    }

    @Override
    protected String doInBackground(String... strings) {
        try {

            List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
            nameValuePairs.add(new BasicNameValuePair("FirstField", FirstValue));
            nameValuePairs.add(new BasicNameValuePair("SecondField", SecondValue));

            HttpClient httpClient = new DefaultHttpClient();
            HttpContext localContext = new BasicHttpContext();

            HttpPost httpPost = new HttpPost(UrlHandler.vRegisterDevice);
            httpPost.setHeader(HTTP.CONTENT_TYPE,"application/x-www-form-urlencoded;charset=UTF-8");
            httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs, "UTF-8"));

            HttpResponse response = httpClient.execute(httpPost);

            BufferedReader reader = new BufferedReader(new InputStreamReader(response.getEntity().getContent(), "UTF-8"));
            String sResponse;
            //Log.e("Response: ", "" + reader.readLine());
            if (sb != null) {
                sb = null;
                sb = new StringBuilder();
            }

            while ((sResponse = reader.readLine()) != null) {
                sb = sb.append(sResponse);
            }

            Log.i("Test","doInBackground: " + sb.toString());
        }catch (Exception e){
            e.printStackTrace();
        }
        return null;
    }

    @Override
    protected void onPostExecute(String s) {
        super.onPostExecute(s);
        try {
            Log.d("Test","onPostExecute - Register Device");
            String vResult = "";
            if(sb != null)
                vResult = sb.toString();

            Log.d("Result String ",vResult);
            if (vResult.contains("Success")) {//success
                Intent intent = new Intent(CurrentAtivity.this,TargetActivity.class);
                startActivity(intent);

                //Hide the progress bar as the task done
                ProgressBar.setVisibility(View.GONE);
            }
        }catch (Exception ex){
            ex.printStackTrace();
        }
    }
}

Answer 2

这是我使用openConnection方法进行Web服务的方式。这是使用get方法，但您可以相应地更改它。

 @Override
    protected Integer doInBackground(String... params) {

        InputStream inputStream = null;

        HttpURLConnection urlConnection = null;

        Integer result = 0;
        try {
            /* forming th java.net.URL object */
            URL url = new URL(params[0]);
            //String mobileNo = mobileno.getText().toString();
            urlConnection = (HttpURLConnection) url.openConnection();

             /* optional request header */
            urlConnection.setRequestProperty("Content-Type", "application/json");

            /* optional request header */
            urlConnection.setRequestProperty("Accept", "application/json");

            /* for Get request */
            urlConnection.setRequestMethod("GET");

            int statusCode = urlConnection.getResponseCode();

            /* 200 represents HTTP OK */
            if (statusCode == 200) {

                inputStream = new BufferedInputStream(urlConnection.getInputStream());

                response = convertInputStreamToString(inputStream);



                result = 1; // Successful

                System.out.println(response);
            } else {
                result = 0; //"Failed to fetch data!";

            }

        } catch (Exception e) {
            Log.d("msg", e.getLocalizedMessage());
        }

        return result; //"Failed to fetch data!";
    }

这是 convertInputStream 方法

private String convertInputStreamToString(InputStream inputStream) throws IOException {

    BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream));

    String line;
    String result = "";

    while ((line = bufferedReader.readLine()) != null) {
        result += line;
    }

    inputStream.close();

    return result;
}

这就是我传递网址的方式

 JOBCOMPLETE_URL=Baseurl + "job_complete.php?cid="+userid;
    CompletedJobAsyncTask completedJobAsyncTask=new CompletedJobAsyncTask();
    completedJobAsyncTask.execute(JOBCOMPLETE_URL);

从Android发送数据到服务器无法正常工作

2 个答案: