Android - JsonException在httppost上

Question

我試圖用我的數據庫連接我的應用程序。

我做這個腳本PHP：

<?php 

/*ci colleghiamo al database(attenti perchè se lavorate in locale 
l'host è 10.0.2.2 e non 127.0.0.1)*/ 
mysql_connect("localhost","mydb","") 
    or die("Impossibile connettersi al server MySQL.\n"); 
//selezioniamo il db a cui ci vogliamo connettere 
mysql_select_db("mydb") 
    or die("Impossibile aprire il database.\n"); 
//creo la tabella nel database 
mysql_query("CREATE TABLE IF NOT EXISTS `mydb`.`survey` (
        `Museum_ID` INT NOT NULL, 
       `Game_ID` INT NOT NULL, 
       `Vote` INT NOT NULL, 
       PRIMARY KEY (`Museum_ID`)) 
       ENGINE = InnoDB;"); 
$idMuseum = $_POST['Museum_ID']; 
$idGame = $_POST['Game_ID']; 
$Vote = $_POST['Vote'] 
$result = mysql_query("INSERT INTO `mydb`.`survey` (`Museum_ID`, `Game_ID`, `Vote`) VALUES ($idMuseum, $idGame, $Vote);"); 
$row = mysqli_fetch_array($result); 
$data = $row[0]; 
if($data){ 
echo $data; 


mysql_close(); 
?> 

，這是我的Java代碼：

public class MakePost extends AsyncTask<String, Void, Void>{ 

    @Override 
     protected void onPreExecute() { 
      super.onPreExecute(); 

      //toggleUI(0); 

      pDialog = new ProgressDialog(Survey.this); 
      pDialog.setMessage("Sending data. Please wait..."); 
      pDialog.setIndeterminate(false); 
      pDialog.setCancelable(false); 
      pDialog.show(); 
     } 

    protected Void doInBackground(String... args) { 

      List<NameValuePair> parametriDaInviare = new ArrayList<NameValuePair>(2); 
      JSONObject json = null; 
      int success = -1; 
      updatesAv = false; 


      try { 
       if(checkedButton == R.id.lowChoise){ 
        //aggiungo alla lista parametri il voto 
        parametriDaInviare.add(new BasicNameValuePair("Vote", "1")); 
        parametriDaInviare.add(new BasicNameValuePair("Museum_ID", "1")); 
        parametriDaInviare.add(new BasicNameValuePair("Game_ID", "1")); 

        // get JSON Object by using POST method 
        json = jParser.makeHttpRequest(url_Survey, "POST", parametriDaInviare);    
        try 
        { 
         Log.d("PHP Response", json.toString()); 
         success = json.getInt(TAG_SUCCESS); 
         if(success == 1) 
         { 
          popupWindow.dismiss(); 
          updatesAv =true; 

         } 
         else { 
          updatesAv = false; 
         } 
         Log.d("UPDATES CHECK", "RESULT: " + updatesAv); 


        }catch (JSONException e) 
        { 
         // TODO Auto-generated catch block 
         e.printStackTrace(); 
         updatesAv = false; 
        } 
       } 
       if(checkedButton == R.id.mediumChoise){ 
        //do something 

        parametriDaInviare.add(new BasicNameValuePair("Vote", "2")); 
        parametriDaInviare.add(new BasicNameValuePair("Museum_ID", "1")); 
        parametriDaInviare.add(new BasicNameValuePair("Game_ID", "1"));      
        // get JSON Object by using POST method 
        json = jParser.makeHttpRequest(url_Survey, "POST", parametriDaInviare);    
        try 
        { 
         Log.d("PHP Response", json.toString()); 
         success = json.getInt(TAG_SUCCESS); 
         if(success == 1) 
         { 
          popupWindow.dismiss(); 
          updatesAv =true; 

         } 
         else { 
          updatesAv = false; 
         } 
         Log.d("UPDATES CHECK", "RESULT: " + updatesAv); 


        }catch (JSONException e) 
        { 
         // TODO Auto-generated catch block 
         e.printStackTrace(); 
         updatesAv = false; 
        } 
       } 
       if(checkedButton == R.id.highChoise){ 
        //do something 

        parametriDaInviare.add(new BasicNameValuePair("Vote", "3")); 
        parametriDaInviare.add(new BasicNameValuePair("Museum_ID", "1")); 
        parametriDaInviare.add(new BasicNameValuePair("Game_ID", "1"));      // get JSON Object by using POST method 
        json = jParser.makeHttpRequest(url_Survey, "POST", parametriDaInviare);    
        try 
        { 
         Log.d("PHP Response", json.toString()); 
         success = json.getInt(TAG_SUCCESS); 
         if(success == 1) 
         { 
          popupWindow.dismiss(); 
          updatesAv =true; 

         } 
         else { 
          updatesAv = false; 
         } 
         Log.d("UPDATES CHECK", "RESULT: " + updatesAv); 


        }catch (JSONException e) 
        { 
         // TODO Auto-generated catch block 
         e.printStackTrace(); 
         updatesAv = false; 
        } 
       } 



      } catch (Exception e) { 
       // TODO Auto-generated catch block 
       // da implementare 
      } 



    return null;  
    } 


    protected void onPostExecute(String file_url) { 
      // dismiss the dialog after getting all data 
      pDialog.dismiss(); 

     } 

當我發送的數據我有此錯誤：

11-03 13:20:55.918: E/JSON Parser(11238): Error parsing data org.json.JSONException: Value <br of type java.lang.String cannot be converted to JSONObject 

我有使用php的經驗並不多，我不知道在我的代碼中是否有一些錯誤。 有人可以幫助我嗎？

感謝

Answer 1

變化
回聲$數據

到 回波json_encode（$數據）;

在該行的末尾添加一個分號

$投票= $ _POST [ '投票'];

Answer 2

您的服務器大多是傳回HTML的應用程序，然後試圖解析失敗。由於錯誤表示響應以br開始。請檢查您是否得到您的服務器的正確響應

Answer 3

您需要更改您的PHP響應爲這樣 echo Json_encode（$ data）; 因爲它發送平面HTML數據，這就是爲什麼你越來越JSONEXCEPTION

Answer 4

1. 您在錯誤的位置在您的查詢分號：

更改此：VALUES ($idMuseum, $idGame, $Vote);");

到VALUES ($idMuseum, $idGame, $Vote)");

1. 正如阿米特庫瑪已經提到，你應該使用echo json_encode($data);

Answer 5

1. 首先你應該刪除$row = mysqli_fetch_array($result); 這是一個插入無法取回！更何況你是混合 的mysqli和mysql擴展
2. 使用的mysqli extenstions，你有機會接觸到SQL注入
3. 最後echo $data;是不是JSON它會破壞你的應用程序，因爲你 atttempt解析JSON數據。
4. 缺少分號$Vote = $_POST['Vote'];

if(!$result){ 
    echo json_encode(array('success'=> 0)); 
}else{ 
    echo json_encode(array('success'=> 1)); 
}