即使结果与 if 语句匹配，Android 也会“else”

Posted 2023-03-31

【中文标题】即使结果与 if 语句匹配，Android 也会“else”

【英文标题】：Android going to 'else' even when result matches the if statement

【发布时间】：2017-05-03 04:54:01

【问题描述】：

我正在开发一个安卓应用程序。 在我的代码的特定部分，总是执行 else 选项，即使满足 if 语句也是如此。非常感谢任何帮助。 这是我的代码：

php：

<?php
error_reporting(0);
include 'conexao.php';

$usuario = $_POST["username"];
$senha   = $_POST["password"];

$result   = mysql_query("select * from login where usuario = '" . $usuario . "' and senha = '" . $senha . "';");
$num_rows = mysql_num_rows($result);

if ($num_rows == 0) 
    echo 'false';
 else if ($num_rows == 1) 
    echo 'true';
 else 
    echo 'nenhum';

?>

还有我的 Java：

EditText txtusuario, txtsenha;
Button btnlogin;
public static final int CONNECTION_TIMEOUT=10000;
public static final int READ_TIMEOUT=15000;

@Override
protected void onCreate(Bundle savedInstanceState) 
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_login);

    txtusuario = (EditText) findViewById(R.id.txtusuario);
    txtsenha = (EditText) findViewById(R.id.txtsenha);
    btnlogin = (Button) findViewById(R.id.btnlogin);

    btnlogin.setOnClickListener(new View.OnClickListener() 
        @Override
        public void onClick(View v) 
            // Get text from email and passord field
            final String username = txtusuario.getText().toString();
            final String password = txtsenha.getText().toString();

            // Initialize  AsyncLogin() class with email and password
            new AsyncLogin().execute(username,password);
        
    );


private class AsyncLogin extends AsyncTask<String, String, String>

    ProgressDialog pdLoading = new ProgressDialog(LoginActivity.this);
    HttpURLConnection conn;
    URL url = null;

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

        //this method will be running on UI thread
        pdLoading.setMessage("\tLoading...");
        pdLoading.setCancelable(false);
        pdLoading.show();

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

            // Enter URL address where your php file resides
            url = new URL("http://rhynotcc.hol.es/teste/login.php");

         catch (MalformedURLException e) 
            // TODO Auto-generated catch block
            e.printStackTrace();
            return "exception";
        
        try 
            // Setup HttpURLConnection class to send and receive data from php and mysql
            conn = (HttpURLConnection)url.openConnection();
            conn.setReadTimeout(READ_TIMEOUT);
            conn.setConnectTimeout(CONNECTION_TIMEOUT);
            conn.setRequestMethod("POST");

            // setDoInput and setDoOutput method depict handling of both send and receive
            conn.setDoInput(true);
            conn.setDoOutput(true);

            // Append parameters to URL
            Uri.Builder builder = new Uri.Builder()
                    .appendQueryParameter("username", params[0])
                    .appendQueryParameter("password", params[1]);
            String query = builder.build().getEncodedQuery();

            // Open connection for sending data
            OutputStream os = conn.getOutputStream();
            BufferedWriter writer = new BufferedWriter(
                    new OutputStreamWriter(os, "UTF-8"));
            writer.write(query);
            writer.flush();
            writer.close();
            os.close();
            conn.connect();

         catch (IOException e1) 
            // TODO Auto-generated catch block
            e1.printStackTrace();
            return "exception";
        

        try 

            int response_code = conn.getResponseCode();

            // Check if successful connection made
            if (response_code == HttpURLConnection.HTTP_OK) 

                // Read data sent from server
                InputStream input = conn.getInputStream();
                BufferedReader reader = new BufferedReader(new InputStreamReader(input));
                StringBuilder result = new StringBuilder();
                String line;

                while ((line = reader.readLine()) != null) 
                    result.append(line);
                

                // Pass data to onPostExecute method
                return(result.toString());

             else 

                return("unsuccessful");
            

         catch (IOException e) 
            e.printStackTrace();
            return "exception";
         finally 
            conn.disconnect();
        
    

    @Override
    protected void onPostExecute(String result) 

        //this method will be running on UI thread
        pdLoading.dismiss();
        String teste = "true";

        if(result.equals(teste)) 
            //Notificação para saber o resultado do login
            int a = 0;
            NotificationCompat.Builder mBuilder = new NotificationCompat.Builder(LoginActivity.this);
            mBuilder.setSmallIcon(R.drawable.escavadeira);
            mBuilder.setContentTitle("Notificação");
            mBuilder.setContentText("Conexão deu certo caralho" + result);
            NotificationManager mNot = (NotificationManager) getSystemService(Context.NOTIFICATION_SERVICE);
            mNot.notify(a, mBuilder.build());
         else 
            int a = 0;
            NotificationCompat.Builder mBuilder = new NotificationCompat.Builder(LoginActivity.this);
            mBuilder.setSmallIcon(R.drawable.escavadeira);
            mBuilder.setContentTitle("Notificação");
            mBuilder.setContentText("Deu errado" + result);
            NotificationManager mNot = (NotificationManager) getSystemService(Context.NOTIFICATION_SERVICE);
            mNot.notify(a, mBuilder.build());
        

    

我的通知中的结果是 'deu errado true' 当 true 应该是 if 语句时，我已经尝试过 result.equals("true")

【问题讨论】：

不知道为什么人们不能帮助新的 SO 成员...放一个日志并打印结果的值。

你能举个小例子吗？

在onPostExecute中，result参数的值是多少？

我解决了，我不知道为什么，但我的回声返回“真”或“假”，开头有一个空格--'。

感谢您的帮助

【参考方案1】：

if ($num_rows == 0) 
    echo 'false';
 else if ($num_rows == 1) 
    echo 'true';
 else 
    echo 'nenhum';

请确保您的代码应该是这样的

if ($num_rows) 
    echo 'false';
 else 
    echo 'nenhum';

【参考方案2】：

@Override
protected void onPostExecute(String result) 
  super.onPostExecute(result);

  //this method will be running on UI thread

  pdLoading.dismiss();
  String teste = "true";

  int a = 0;
  NotificationCompat.Builder mBuilder = new
  NotificationCompat.Builder(LoginActivity.this);
  mBuilder.setSmallIcon(R.drawable.escavadeira);
  mBuilder.setContentTitle("Notificação");

  if (result.equals(teste)) 
    //Notificação para saber o resultado do login
    mBuilder.setContentText("Conexão deu certo caralho" + result);
   else 
    mBuilder.setContentText("Deu errado" + result);
  

  NotificationManager mNot = (NotificationManager) getSystemService(Context.NOTIFICATION_SERVICE);
  mNot.notify(a, mBuilder.build());

【讨论】：

仍然显示'deu errado true'，这是else语句：/