【问题标题】:Nothing happens when jsonResponse gets null data from mySQL-DB当 jsonResponse 从 mySQL-DB 获取空数据时没有任何反应
【发布时间】:2018-02-14 13:50:06
【问题描述】:

我刚接触 java/android 编程。

我正在编写一个应用程序,用户可以在其中注册并登录。数据保存在在线 mysql-db 中。注册和登录工作正常。用户使用会话保持登录状态。

即使从 mysql-db 获取数据也可以,但是当某些 db 字段响应“null”时会出现一个问题。

这是我正在使用的代码

    public class UserProfileSettingsFragment extends PreferenceFragment
{

    SessionManager session;

    @Override
    public void onCreate(final Bundle savedInstanceState)
    {
        SharedPreferences prefs = this.getActivity().getSharedPreferences("JampSharedPrefs", Context.MODE_PRIVATE);
        SharedPreferences.Editor editor = prefs.edit();

        super.onCreate(savedInstanceState);
        addPreferencesFromResource(R.xml.usersettings);

        session = new SessionManager(this.getActivity().getApplicationContext());



        HashMap<String,String> user = session.getUserDetails();
        final String sessionUsername = user.get(SessionManager.KEY_USERNAME);

        // ResponseListener um Request Nutzerdaten auszulesen.
        Response.Listener<String> UserDataResponseListener = new Response.Listener<String>(){
            @Override
            public void onResponse(String response) {
                try {
                    JSONObject jsonResponse = new JSONObject(response);
                    boolean success = jsonResponse.getBoolean("success");

                    // Wenn Datenabfrage erfolgreich, JSONResponse auswerten.
                    if (success) {
                        String responseRealName   = jsonResponse.getString("realname");
                        String responseStreetName = jsonResponse.getString("streetname");
                        int    responsePostcode   = jsonResponse.getInt   ("postcode");
                        String responseCity       = jsonResponse.getString("city");
                        String responseState      = jsonResponse.getString("state");
                        int    responseAge        = jsonResponse.getInt   ("age");
                        int    responseIsPremium  = jsonResponse.getInt   ("isPremium"); // BOOLEAN

                        Preference prefUserData = (Preference) findPreference("preferencescreen_userdata");
                        prefUserData.setTitle(sessionUsername);
                        //prefUserData.setSummary(responseRealName+"\n"+responseStreetName+"\n"+responsePostcode + " " + responseCity);

                        Preference prefUsername = (Preference) findPreference("settings_username");
                        prefUsername.setTitle(sessionUsername);

                        Toast.makeText(getActivity(),sessionUsername, Toast.LENGTH_LONG);

                        if (responseIsPremium==1){
                            //ivPremiumIcon.setVisibility(View.VISIBLE);
                        }


                    }else{
                        AlertDialog.Builder builder = new AlertDialog.Builder(getActivity());
                        builder.setMessage("Konnte Nutzerdaten nicht abrufen.")
                                .setNegativeButton("Nochmal",null)
                                .create()
                                .show();
                    }

                } catch (JSONException e) {
                    e.printStackTrace();
                }

            }


        };

        // Request an userdatarequest.php senden
        UserDataRequest userDataRequest = new UserDataRequest(sessionUsername, UserDataResponseListener);
        RequestQueue queue = Volley.newRequestQueue(this.getActivity());
        queue.add(userDataRequest);


    }
}

PHP 代码:

$con = mysqli_connect("localhost","web506","lalala","usr_web506_1");

$username = $_POST["username"];

$statement = mysqli_prepare($con,"SELECT * FROM user WHERE username = ?");

mysqli_stmt_bind_param($statement,"s",$username);
mysqli_stmt_execute($statement);

mysqli_stmt_store_result($statement);


mysqli_stmt_bind_result($statement, 
                        $userID, 
                        $username, 
                        $password, 
                        $email, 
                        $age, 
                        $realname, 
                        $streetname, 
                        $postcode, 
                        $city,
                        $state, 
                        $isPremium, 
                        $isLoggedIn);

$response = array();
$response["success"] = false;

while(mysqli_stmt_fetch($statement)){
    $response["success"] = true;        
    $response["username"] = $username;
    $response["password"] = $password;
    $response["email"] = $email;
    $response["age"] = $age;
    $response["realname"] = $realname;
    $response["streetname"] = $streetname;
    $response["postcode"] = $postcode;
    $response["city"] = $city;
    $response["state"] = $state;
    $response["isPremium"] = $isPremium;
    $response["isLoggedIn"] = $isLoggedIn;

}

echo json_encode($response);

?>

所以,当我获取用户数据时,我可以用 Toast 显示它们,更改preference.summaries 或其他任何东西。但是,如果某些 mysql 条目为空/null,则什么也不会发生。应用程序没有崩溃,但似乎它没有从 php 文件中获得“成功”布尔值。什么线索?

提前致谢。 埃里克


我应该删除$response["success"] = false; 吗? 通常,如果应用程序无法连接到数据库并且 false bool 到达我的应用程序,我会收到一条警报消息,所以我认为它就在那里。

当我在我知道它们的 DB 单元为空的变量后面添加空格时,jsonresponse 会提供一个“0”值作为字符串结果,如下所示:

      $response["realname"] = $realname+" ";
      $response["streetname"] = $streetname+" ";
      $response["postcode"] = $postcode+" ";
      $response["city"] = $city+" ";
      $response["state"] = $state+ " ";

我是一个文本视图,它们是否逐行显示为“0”。

我必须在我的应用程序中解决这个问题,还是有一种简单的方法可以以某种方式过滤空单元格并跳到下一个?

【问题讨论】:

  • 因为如果没有数据,您已将其设置为 false $response["success"] = false;

标签: java php android mysql json


【解决方案1】:
    $response["success"] = true;
$record_size = 0;

while(mysqli_stmt_fetch($statement)){
    $response["success"] = true;        
    $response["username"] = $username;
    $response["password"] = $password;
    $response["email"] = $email;
    $response["age"] = $age;
    $response["realname"] = $realname;
    $response["streetname"] = $streetname;
    $response["postcode"] = $postcode;
    $response["city"] = $city;
    $response["state"] = $state;
    $response["isPremium"] = $isPremium;
    $response["isLoggedIn"] = $isLoggedIn;
    $record_size++;
}

$response["record_size"]  = $record_size;

echo json_encode($response);

对于记录数,我使用 $record_size 变量,以便您了解记录。因为 $response["success"] = true;意味着您可以成功获取数据库,并且可以使用 $response["record_size"].. 希望对您有所帮助..

【讨论】:

    【解决方案2】:

    我的解决方法现在看起来像这样: 如果我得到 0 个值,我将它们替换为 xml 资源中的“缺少 [...]”。

    还有另一个不匹配。似乎我总是将我的 jsonresponse 作为字符串,但我想在我的第一个代码中将邮政编码作为整数,这也不起作用。所以当我需要这种方式时,我必须将它解析为一个 int。

    String responseRealName   = jsonResponse.getString("realname");     if (responseRealName.equals("0"))   {responseRealName = getResources().getString(R.string.MissingRealName);}
                            String responseStreetName = jsonResponse.getString("streetname");   if (responseStreetName.equals("0")) {responseStreetName = getResources().getString(R.string.MissingStreetName);}
                            String responsePostcode   = jsonResponse.getString("postcode");     if (responsePostcode.equals("0"))   {responsePostcode = getResources().getString(R.string.MissingPostcode);}
                            String responseCity       = jsonResponse.getString("city");         if (responseCity.equals("0"))       {responseCity = getResources().getString(R.string.MissingCity);}
                            String responseState      = jsonResponse.getString("state");        if (responseState.equals("0"))      {responseState = getResources().getString(R.string.MissingState);}
    

    我认为我的问题已经得到了足够的回答。

    【讨论】:

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