Question

在Android上的AsyncTask中对HTTPURLCONNECTION进行POST的最简单方法是什么？我只想在我的php文件上发布数据，然后返回一个json响应

Answer 1

你可以从我前几天给你的答案中把它拼凑起来： How to get JSON object using HttpURLConnection instead of Volley?

......答案在这里：Java - sending HTTP parameters via POST method easily

话虽这么说，开始发送POST数据并获得JSON结果的最简单方法是使用旧的API。

这是一个有效的例子：

class CreateNewProduct extends AsyncTask<String, String, JSONObject> {

        InputStream is = null;
        JSONObject jObj = null;
        String json = "";

        /**
         * Before starting background thread Show Progress Dialog
         * */
        @Override
        protected void onPreExecute() {
            super.onPreExecute();
            pDialog = new ProgressDialog(ShareNewMessage.this);
            pDialog.setMessage("Sharing Message...");
            pDialog.setIndeterminate(false);
            pDialog.setCancelable(true);
            pDialog.show();
        }

        /**
         * Creating product
         * */
        protected JSONObject doInBackground(String... args) {

            String message = inputMessage.getText().toString();

            // Building Parameters
            List<NameValuePair> params = new ArrayList<NameValuePair>();
            params.add(new BasicNameValuePair("name", username));
            params.add(new BasicNameValuePair("message", message));

            Log.d("Share", "Sharing message, username: " + username + " message: " + message);

            try {
                DefaultHttpClient httpClient = new DefaultHttpClient();
                HttpPost httpPost = new HttpPost(url_create_message);
                httpPost.setEntity(new UrlEncodedFormEntity(params));

                HttpResponse httpResponse = httpClient.execute(httpPost);
                HttpEntity httpEntity = httpResponse.getEntity();
                is = httpEntity.getContent();

            } catch (UnsupportedEncodingException e) {
                e.printStackTrace();
            } catch (ClientProtocolException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }


            try {
                BufferedReader reader = new BufferedReader(new InputStreamReader(
                        is, "iso-8859-1"), 8);
                StringBuilder sb = new StringBuilder();
                String line = null;
                while ((line = reader.readLine()) != null) {
                    sb.append(line + "\n");
                }
                is.close();
                json = sb.toString();
            } catch (Exception e) {
                Log.e("Share", "Error converting InputStream result " + e.toString());
            }

            // try parse the string to a JSON object
            try {
                jObj = new JSONObject(json);
            } catch (JSONException e) {
                Log.e("Share", "Error parsing JSON data " + e.toString());
            }

            try{
                int success = jObj.getInt(TAG_SUCCESS);

                if (success == 1) {
                    // successfully created product

                } else {
                    // failed to create product
                }
            } catch (JSONException e) {
                e.printStackTrace();
            }

            return jObj;
        }

        /**
         * After completing background task Dismiss the progress dialog
         * **/
        protected void onPostExecute(JSONObject jObj) {

            //do something with jObj

            // dismiss the dialog once done
            pDialog.dismiss();
        }

    }

Answer 2

以下是如何将图像上传到服务器的示例代码

class ImageUploadTask extends AsyncTask <Void, Void, String>{
    @Override
    protected String doInBackground(Void... unsued) {
        try {
            HttpClient httpClient = new DefaultHttpClient();
            HttpContext localContext = new BasicHttpContext();
            HttpPost httpPost = new HttpPost(
                    getString(R.string.WebServiceURL)
                            + "/cfc/iphonewebservice.cfc?method=uploadPhoto");

            MultipartEntity entity = new MultipartEntity(
                    HttpMultipartMode.BROWSER_COMPATIBLE);

            ByteArrayOutputStream bos = new ByteArrayOutputStream();
            bitmap.compress(CompressFormat.JPEG, 100, bos);
            byte[] data = bos.toByteArray();
            entity.addPart("photoId", new StringBody(getIntent()
                    .getStringExtra("photoId")));
            entity.addPart("returnformat", new StringBody("json"));
            entity.addPart("uploaded", new ByteArrayBody(data,
                    "myImage.jpg"));
            entity.addPart("photoCaption", new StringBody(caption.getText()
                    .toString()));
            httpPost.setEntity(entity);
            HttpResponse response = httpClient.execute(httpPost,
                    localContext);
            BufferedReader reader = new BufferedReader(
                    new InputStreamReader(
                            response.getEntity().getContent(), "UTF-8"));

            String sResponse = reader.readLine();
            return sResponse;
        } catch (Exception e) {
            if (dialog.isShowing())
                dialog.dismiss();
            Toast.makeText(getApplicationContext(),
                    getString(R.string.exception_message),
                    Toast.LENGTH_LONG).show();
            Log.e(e.getClass().getName(), e.getMessage(), e);
            return null;
        }

        // (null);
    }

有关此问题的完整教程，请浏览this。

在上面的代码中，您只需添加

即可发送要发送的任何表单数据

entity.addPart("value", new StringBody(key));

Answer 3

试一下

    public class parseProduct extends AsyncTask<Void, Void, Void> {
String jsonResponse;
        @Override
        protected void onPreExecute() {

            super.onPreExecute();
        }

        @Override
        protected Void doInBackground(Void... params) {
            JSONObject object = new JSONObject();
            try {
                object.put("registeruserid", userId); // you can pass data which you want to pass from POST
            } catch (JSONException e) {
                e.printStackTrace();
            }
            try {
                jsonResponse=getResponseStringFromURL2(URL,object.toString());

            } catch (ClientProtocolException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            } catch (JSONException e) {
                e.printStackTrace();
            }
            return null;
        }

        @Override
        protected void onPostExecute(Void resul) {
            super.onPostExecute(resul);


        }

    }

以下是功能：

public String getResponseStringFromURL2(String url, String json)
            throws ClientProtocolException, IOException {
        StringBuilder result = new StringBuilder();
        HttpParams httpParameters = new BasicHttpParams();
        int timeoutConnection = 30000;
        HttpConnectionParams.setConnectionTimeout(httpParameters,
                timeoutConnection);
        int timeoutSocket = 30000;
        HttpConnectionParams.setSoTimeout(httpParameters, timeoutSocket);
        DefaultHttpClient httpClient = new DefaultHttpClient(httpParameters);
        HttpPost request = new HttpPost(url);
        if (json != null) {
            StringEntity se = new StringEntity(json);
            request.setEntity(se);
        }

        request.setHeader("Accept", "application/json");
        request.setHeader("Content-type", "application/json");
        Log.d("request: ", ":" + request.getRequestLine());
        HttpResponse response = null;
        response = httpClient.execute(request);
        if (response == null)
            return null;

        InputStream input = null;

        input = new BufferedInputStream(response.getEntity().getContent());

        byte data[] = new byte[40000];

        int currentByteReadCount = 0;

        /** read response from inpus stream */

        while ((currentByteReadCount = input.read(data)) != -1) {
            String readData = new String(data, 0, currentByteReadCount);
            result.append(readData);
        }

        input.close();

        return result.toString();

    }

在HTTPURLConnection上进行POST的简便方法？

3 个答案: