Получение ответа от вызова php-скрипта

Question

У меня есть приложение, которое отправляет некоторую информацию через POST скрипту php на сервере. Он использует Asynchttpclient. Как я могу получить ответ с сервера (через json?)? Пожалуйста помоги.

Это мой скрипт

if($_POST["mode"]=="newuser"){ 
      //$gcmRegID = $_GET["shareRegId"]; 
      $gcmRegID = $_POST["regID"]; 
      $gcmUserName = $_POST["userName"]; 
      $gcmFolderName = $_POST["folderName"]; 

      $gcmDate = date("d/m/y"); 

      $conn = new mysqli($servername, $username, $password, $dbname); 
      if($conn->connect_error){ 
       die("Connection failed: " . $conn->connect_error); 
      } 
      $in_user = "user"; 
      $in_password = "NULL"; 
      $in_email = "NULL"; 
      $in_dob = "NULL"; 
      $in_role = "user"; 
      $in_datejoined = "0000-00-00"; 
      $foldername = "NULL"; 

      $sql = "INSERT INTO user(password,regid,name,email,phone,dob,role,datejoined,foldername) VALUES('$in_password','$gcmRegID','$gcmUserName','$in_email','$in_phone','$in_dob','$in_role','$gcmDate','$foldername')"; 

      $substringtitle = substr($gcmRegID,-7); 
      $combined = $gcmUserName."_".$substringtitle; 
      if($conn->query($sql)===TRUE){ 
        mkdir("./users/".$gcmFolderName); 
        $newfoldername = "./users/".$gcmFolderName; 
        $updatequery = "UPDATE user SET foldername='$newfoldername' WHERE name='$gcmUserName'"; 


        $returnfield = array(
         'foldername' => $newfoldername 
        ); 
        header('Content-type: application/json'); 
        echo json_encode(array('returnfield'=>$returnfield)); 


        if($conn->query($updatequery)===TRUE){ 
         echo "folder updated"; 
        } 
        //echo "Folder created!"; 
       //} 
       echo "New record created successfully"; 

      }else{ 
       echo "Error: " . $sql . "<br>" . $conn->error; 
      } 
      $conn->close(); 

      echo "Done!"; 
      exit; 
     } 

Android код

//store in the file server (PHP) 
private void storeREG(final String registerID,String userName,String folderName){ 

    pg.show(); 
    params.put("regID", registerID); 
    params.put("userName",userName); 
    params.put("folderName", folderName); 
    params.put("mode","newuser"); 
    Log.d("STORE","STORE"); 
    //Make RESTful webservice call 

    AsyncHttpClient client = new AsyncHttpClient(); 
    client.post(AppConstants.SERVER_URL, params, new AsyncHttpResponseHandler() { 

     @Override 
     public void onSuccess(String content) { 
      pg.hide(); 
      if (pg != null) { 
       pg.dismiss(); 
      } 


      Toast.makeText(applicationCtx, "ID sharing successful", Toast.LENGTH_LONG).show(); 
      Intent home = new Intent(applicationCtx, HomeActivity.class); 
      home.putExtra("regID", registerID); 
      Log.d("REGID", registerID); 
      startActivity(home); 
      finish(); 
     } 

     @Override 
     public void onFailure(int statusCode, Throwable error, String content) { 
      pg.hide(); 
      if (pg != null) { 
       pg.dismiss(); 
      } 
      Log.d("ERRORTHROW", error.toString()); 
      if (statusCode == 404) { 
       Toast.makeText(applicationCtx, "Requested resource not found", Toast.LENGTH_LONG).show(); 
      } else if (statusCode == 500) { 
       Toast.makeText(applicationCtx, "Something went wrong at the server", Toast.LENGTH_LONG).show(); 
      } else { 
       Log.d("SHOWME", String.valueOf(statusCode)); 
       Toast.makeText(applicationCtx, "Unexpected error occurred", Toast.LENGTH_LONG).show(); 
      } 
     } 


    }); 

} 

Надеюсь, я могу получить помощь с этим.

Answer 1

Вы можете попытаться пересмотреть свой PHP-код. Ниже приведено хорошо прокомментированный образец кода, чтобы вы начали:

<?php 
     // EXPLICITLY INSTRUCT THE HEADER ABOUT THE CONTENT TYPE. HERE - JSON 
     header('Content-type: application/json'); 

     if($_POST["mode"]=="newuser"){ 
      $gcmRegID  = htmlspecialchars(trim($_POST["regID"])); 
      $gcmUserName = htmlspecialchars(trim($_POST["userName"])); 
      $gcmFolderName = htmlspecialchars(trim($_POST["folderName"])); 
      $gcmDate  = date("d/m/y"); 

      // I WOULD STRONGLY SUGGEST YOU USE PDO FOR YOUR DATABASE TRANSACTIONS: 
      // HERE'S HOW: 

      //DATABASE CONNECTION CONFIGURATION: 
      defined("HOST")  or define("HOST", "localhost");   //REPLACE WITH YOUR DB-HOST 
      defined("DBASE") or define("DBASE", "database");   //REPLACE WITH YOUR DB NAME 
      defined("USER")  or define("USER", "root");    //REPLACE WITH YOUR DB-USER 
      defined("PASS")  or define("PASS", "root");    //REPLACE WITH YOUR DB-PASS 

      // ESTABLISH A CONNECTION AND DO YOUR WORK WITHIN A TRY-CATCH BLOCK... 
      try { 
       $dbh  = new PDO('mysql:host='.HOST.';dbname='. DBASE,USER,PASS); 
       $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 

       // HERE: ALL YOUR BUSINESS LOGIC... 
       $in_user  = "user"; 
       $in_password = "NULL"; 
       $in_email  = "NULL"; 
       $in_phone  = "NULL"; 
       $in_dob   = "NULL"; 
       $in_role  = "user"; 
       $in_dateJoined = "0000-00-00"; 
       $folderName  = "NULL"; 

       $insertSQL  = "INSERT INTO user(`password`, `regid`, `name`, `email`, `phone`, `dob`, `role`, `datejoined`, `foldername`) "; 
       $insertSQL  .= " VALUES(:inPassword, :gcmRegID, :gcmUserName, :inEmail, :inPhone, :inDOB, :inRole, :gcmDate, :folderName)"; 

       $arrInsertData = array(
        'inPassword' => $in_password, 
        'gcmRegID'  => $gcmRegID, 
        'gcmUserName' => $gcmUserName, 
        'inEmail'  => $in_email, 
        'inPhone'  => $in_phone, 
        'inDOB'   => $in_dob, 
        'inRole'  => $in_role, 
        'gcmDate'  => $gcmDate, 
        'folderName' => $folderName 
       ); 

       // PREPARE THE INSERT QUERY: 
       $insertStmt = $dbh->prepare($insertSQL); 

       // INSERT THE NEW ROW: 
       $insertStmt->execute($arrInsertData); 

       // OBTAIN THE ID OF THE INSERTED ROW TO BE USED AS SUFFIX FOR YOUR USER FOLDER 
       $id   = $dbh->lastInsertId(); 

       // WHAT HAPPENS WHEN 2 USERS HAVE THE SAME USERNAME??? DID YOU THINK ABOUT THAT? 
       // TO CIRCUMVENT THIS ISSUE; I WOULD SUGGEST FIRST TO INSERT THE DATA TO THE DATABASE... 
       // THEN USE THE ID AS A SUFFIX TO MAKE EACH USER DIRECTORY UNIQUE & THAT IS THE APPROACH TAKEN HERE THOUGH... 

       // NOW YOU CAN CREATE YOUR FOLDER USING THIS ID: $id 
       // LIKE THIS; 2 USERS WITH USERNAME "android_user" CAN HAVE 2 DIFFERENT FOLDERS LIKE SO: "android_user_97" & "android_user_102" 
       $userDirectory = "./users/" . $gcmFolderName . "_" . $id; 
       mkdir($userDirectory); 

       // DID IT OCCUR TO YOU THAT 2 USERS MIGHT HAVE THE SAME USERNAME IN WHICH CASE MYSQL (INSTEAD OF YOU) HAS TO DECIDE WHICH USER TO UPDATE? 
       // THAT IS WHY DATABASE TABLES ARE DESIGNED TO HAVE UNIQUE IDENTIFIERS LIKE UUID OR ID OR UID OR ANY TOKEN TO MAKE EACH ROW UNIQUE... 
       // WE ARE ADOPTING THIS APPROACH IN THE UPDATE QUERY... THAT IS: WE UPDATE THE ROW USING THE ID ABOVE... ASSUMING THAT IS A UNIQUE COLUMN THOUGH. 
       $updateSQL  = "UPDATE user SET foldername=:newDirName WHERE id=:ID"; 

       // NOW UPDATE THE ROW TO TAKE INTO ACCOUNT THE UNIQUE USER-DIRECTORY (USING THE ID AS THE KEY) 
       $arrUpdateData = array(
        'newDirName' => $userDirectory, 
        'ID'   => $id  // THIS ASSUMES THAT THE PRIMARY KEY OF YOUR TABLE IS CALLED id OTHERWISE USE THE APPROPRIATE KEY NAME: EG: reg_id OR WHATEVER 
       ); 

       // PREPARE THE UPDATE QUERY: 
       $insertStmt = $dbh->prepare($updateSQL); 

       // UPDATE THE NEWLY CREATED ROW: 
       $insertStmt->execute($arrUpdateData); 

       // BUILD THE RESPONSE JSON DATA 
       $arrResponse = array(
        'folderName' => $userDirectory, 
        'id'   => $id, 
       ); 

       // SEND THE RESPONSE AS JSON IF ALL WORKS FINE TILL HERE... 
       // THAT MEANS: SEND THE DATA IN $arrResponse AND TERMINATE THE SCRIPT - THE JOB IS DONE. 
       // NO NEED FOR ALL THOSE ECHO STATEMENTS AS THE YOU ARE EXPLICITLY SENDING BACK JSON DATA. 
       die(json_encode($arrResponse)); 

      }catch(PDOException $e){ 
       // IF THERE WAS ANY KIND OF PDO ERROR, SEND IT BACK ANYWAYS - BUT ALSO AS JSON:    
       $arrResponse = array(
        'error'  => $e->getMessage() 
       ); 
       die(json_encode($arrResponse)); 
      } 
     } 

Answer 2

Я не могу проверить код прямо сейчас (извините), но я думаю, что это должно быть что-то вроде этого:

try { 
    RequestParams rParams = new RequestParams(); 
    rParams.put("example", "example"); // POST 
    AsyncHttpClient client = new AsyncHttpClient(); 
    client.get(pageURL, rParams, new JsonHttpResponseHandler() { 

     @Override 
     public void onSuccess(JSONArray jsonArray) { 
      super.onSuccess(jsonArray); 

      //process JSON Array 

     } 

     @Override 
     public void onFailure(Throwable throwable, JSONArray jsonArray) { 
      super.onFailure(throwable, jsonArray); 
      Log.d(TAG, "error", throwable); 
     } 

    }); 
} catch (Exception e) { 
    Log.d(TAG, "exception", e); 
} 

В противном случае я сделал очень легкий WebClient, вы можете дать ему шанс :

https://github.com/omaflak/WebClient

Это короткий пример:

WebClient client = new WebClient(); 
client.setOnRequestListener(new OnRequestListener() { 
     @Override 
     public void onRequest(String response, int requestID) { 
      Log.e(TAG, response); 
     } 

     @Override 
     public void onError(int error_code, String message) { 
      Log.e(TAG, message); 
     } 
}); 

Pair p = new Pair("field1", "value1"); 
Pair p2 = new Pair("field2", "value2"); 
client.requestAsync("http://your-api.com", WebClient.POST, Arrays.asList(p, p2), 1); 

// requestAsync(String url, String method, List<Pair<String, String>> postData, int requestID) 

Чтобы использовать его, просто добавьте к вашим зависимостям:

compile 'me.aflak.libraries:webclient:1.0'