Question

从第一张图片中我发送请求的两项活动到两个不同的PHP php文件，我得到了有效的Android活动响应。

在第二张图片中，将上述两个php文件合并为一个 php文件，当从Activity 1发送请求到php时，我无法做到至得到回应和活动2。

 private void login(final String textViewNameq, String textViewNameqw,String day) {

    class LoginAsync extends AsyncTask<String, Void, String> {

        private Dialog loadingDialog;

        @Override
        protected void onPreExecute() {
            super.onPreExecute();
            loadingDialog = ProgressDialog.show(MainActivity.this, "Please wait", "Loading...");
        }

        @Override
        protected String doInBackground(String... params) {
            String uname = params[0];
            String pass = params[1];
            String day=params[2];

            InputStream is = null;
            List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
            nameValuePairs.add(new BasicNameValuePair("username", uname));
            nameValuePairs.add(new BasicNameValuePair("password", pass));
            nameValuePairs.add(new BasicNameValuePair("day", day));
            String result = null;

            try {
                HttpClient httpClient = new DefaultHttpClient();
                HttpPost httpPost = new HttpPost(
                        "Your php link goes here");
                httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

                HttpResponse response = httpClient.execute(httpPost);

                HttpEntity entity = response.getEntity();

                is = entity.getContent();

                BufferedReader reader = new BufferedReader(new InputStreamReader(is, "UTF-8"), 8);
                StringBuilder sb = new StringBuilder();

                String line = null;
                while ((line = reader.readLine()) != null) {
                    sb.append(line + "\n");
                }
                result = sb.toString();
            } catch (ClientProtocolException e) {
                e.printStackTrace();
            } catch (UnsupportedEncodingException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return result;
        }

和我的PHP代码在这里，

   <?php
include("my php file path");                        
//  if (!empty($_POST['username']) && !empty($_POST['password'])){
        $Fromname = trim($_POST['username']);
        $Toname = trim($_POST['password']);
        //$Fromdate=$_POST['iddate'];
        //$Fromd = date("Y-m-d", strtotime($Fromdate));
        $from_id=trim($_POST['idss']);
        $to_id=trim($_POST['iddd']);
    //  }

     $url="http://my url with API key goes here";


     $headers = array('Content-Type: application/json');                
            $ch = curl_init();
                curl_setopt($ch, CURLOPT_URL, $url);
                curl_setopt($ch, CURLOPT_POST, true);   
                curl_setopt($ch, CURLOPT_HTTPHEADER, $headers);                 
                curl_setopt($ch, CURLOPT_RETURNTRANSFER, true );
                curl_setopt($ch, CURLOPT_ENCODING , "gzip");
                curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);                        
                $page = curl_exec($ch);
                curl_close($ch);    
                $result=json_decode($page,true);     
                if(empty($result)){                 
                    $page=file_get_contents($url);
                    $result=json_decode($page,true);
                }

Answer 1

可以从同一个php文件中获取所需的响应。您需要做的就是在请求方法中发送所需信息并在php文件中进行相应处理并生成响应。如果你展示你的Php脚本，我们可以帮助你更好。

如何将数据从活动发送到php并从同一个php检索数据到另一个活动？

1 个答案: