Android的Mysql連接錯誤

Question

嘿，夥計們我試圖使註冊頁面使用mysqli和PHP的後端和Java的前端。我的PHP連接文件工作文件，即使我插入數據以及當我試圖插入數據使用Android應用程序它不插入，甚至它不顯示任何錯誤日誌。

我是在android的Begginer。請原諒，如果你發現任何錯誤。

這裏是我的插入文件

require_once('dbConnect.php'); 


$email = $_POST['email']; 
$password = $_POST['password']; 
$full_name = $_POST['full_name']; 
$registration_num = $_POST['registration_num']; 
$contact_num = $_POST['contact_num']; 



$sql = "INSERT INTO tblregistration (Email, Password,Full_name,Registration_number,Contact_number) VALUES ($email,$password,$full_name,$registration_num,$contact_num)"; 

if(mysqli_query($con,$sql)) { 
    echo "Successful"; 
} 

mysqli_close($con); 

這是我的註冊頁面

b1 = (Button)findViewById(R.id.button); 
    b1.setOnClickListener(new View.OnClickListener() { 
     @Override 
     public void onClick(View v) { 

      str_email = "a"; 
      str_password = "a"; 
      str_full_name = "a"; 
      str_registration_num = "a"; 
      str_contact_number = "a"; 
      new InsertData().execute(); 
     } 
    }); 

private class InsertData extends AsyncTask<String, Void, String> { 

    String JSON_STRING; 
    JSONObject jsonObject = null; 

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

    @Override 
    protected void onPostExecute(String s) { 
     super.onPostExecute(s); 

      Toast.makeText(getApplicationContext(), s, Toast.LENGTH_SHORT).show(); 

    } 

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

     JSONParser rh = new JSONParser(); 

     HashMap<String, String> data = new HashMap<>(); 

     data.put("email", str_email); 
     data.put("password", str_password); 
     data.put("full_name", str_full_name); 
     data.put("registration_num", str_registration_num); 
     data.put("contact_num", str_contact_number); 

     JSON_STRING = rh.sendPostRequest(AppConfig.URL_REGISTER, data); 

     return JSON_STRING; 
    } 
} 

這裏是我的HttpURLConnection的文件

public String sendPostRequest(String requestURL, 
           HashMap<String, String> postDataParams) { 
    //Creating a URL 
    URL url; 

    //StringBuilder object to store the message retrieved from the server 
    StringBuilder sb = new StringBuilder(); 
    try { 
     //Initializing Url 
     url = new URL(requestURL); 

     //Creating an httmlurl connection 
     HttpURLConnection conn = (HttpURLConnection) url.openConnection(); 

     //Configuring connection properties 
     conn.setReadTimeout(15000); 
     conn.setConnectTimeout(15000); 
     conn.setRequestMethod("POST"); 
     conn.setDoInput(true); 
     conn.setDoOutput(true); 

     //Creating an output stream 
     OutputStream os = conn.getOutputStream(); 

     //Writing parameters to the request 
     //We are using a method getPostDataString which is defined below 
     BufferedWriter writer = new BufferedWriter(
       new OutputStreamWriter(os, "UTF-8")); 
     writer.write(getPostDataString(postDataParams)); 

     writer.flush(); 
     writer.close(); 
     os.close(); 
     int responseCode = conn.getResponseCode(); 

     if (responseCode == HttpsURLConnection.HTTP_OK) { 

      BufferedReader br = new BufferedReader(new InputStreamReader(conn.getInputStream())); 
      sb = new StringBuilder(); 
      String response; 
      //Reading server response 
      while ((response = br.readLine()) != null){ 
       sb.append(response); 
      } 
     } 

    } catch (Exception e) { 
     e.printStackTrace(); 
    } 
    return sb.toString(); 
} 


private String getPostDataString(HashMap<String, String> params) throws UnsupportedEncodingException { 
    StringBuilder result = new StringBuilder(); 
    boolean first = true; 
    for (Map.Entry<String, String> entry : params.entrySet()) { 
     if (first) 
      first = false; 
     else 
      result.append("&"); 

     result.append(URLEncoder.encode(entry.getKey(), "UTF-8")); 
     result.append("="); 
     result.append(URLEncoder.encode(entry.getValue(), "UTF-8")); 
    } 

    return result.toString(); 
} 

Answer 1

一切都看起來不錯。使用適當的JSON格式... !!!