Question

我正在创建一个Android应用程序，将打开的第一个活动将是一个密码输入。要知道密码已经在变量中设置了值“admit”。我创建了一个if语句来检查用户输入了什么值，然后将其与变量进行比较，如果值匹配则转到主屏幕，但是它一直说它是错误的密码。任何人都可以查看我的代码，并建议我出错了。

public class Password extends ActionBarActivity implements View.OnClickListener {
    protected String password = "admin";
    String getPassword;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_password);
        Button passwordButton = (Button) findViewById(passwordbutton);
        EditText passwordInput = (EditText) findViewById(R.id.password);
        getPassword = (passwordInput.getText().toString());
        passwordButton.setOnClickListener(this);

    }

    public void onClick(View v) {

        if (getPassword.equals(password)) {
            Intent goHome;
            goHome = new Intent(this, home.class);
            startActivity(goHome);
        } else {
             AlertDialog.Builder wrongPasswordBuilder = new AlertDialog.Builder(this);
            wrongPasswordBuilder.setTitle(getString(R.string.wrongPasswordTitle));
            wrongPasswordBuilder.setMessage(getString(R.string.wrongPasswordTryAgain));
            wrongPasswordBuilder.setPositiveButton("ok", null);
            AlertDialog dialog = wrongPasswordBuilder.show();
        }
    }

Answer 1

如果在onCreate（）中写入passwordInput.getText（）。toString（），它总是返回空字符串。所以在onClick（）中写下这些行。代码写在下面

public void onClick(View v) {
getPassword = passwordInput.getText().toString()
if (getPassword.equals(password)) {
    Intent goHome;
    goHome = new Intent(this, home.class);
    startActivity(goHome);
} else {
     AlertDialog.Builder wrongPasswordBuilder = new AlertDialog.Builder(this);
    wrongPasswordBuilder.setTitle(getString(R.string.wrongPasswordTitle));
    wrongPasswordBuilder.setMessage(getString(R.string.wrongPasswordTryAgain));
    wrongPasswordBuilder.setPositiveButton("ok", null);
    AlertDialog dialog = wrongPasswordBuilder.show();
       }
}

Answer 2

将getPassword = passwordInput.getText().toString();放在OnClickListener中。现在，您的变量getPassword正在onCreate方法中设置，因此它将始终设置为EditText的默认值。相反，您需要在每次按下按钮时更新变量。

Answer 3

@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_password);
Button passwordButton = (Button) findViewById(passwordbutton);
EditText passwordInput = (EditText) findViewById(R.id.password);

passwordButton.setOnClickListener(this);

}

public void onClick(View v) {
//Place the getPassword variable in the onClick method. That way it's stored everytime you click it.
getPassword = (passwordInput.getText().toString());

if (getPassword.equals(password)) {
    Intent goHome;
    goHome = new Intent(this, home.class);
    startActivity(goHome);
} else {
     AlertDialog.Builder wrongPasswordBuilder = new AlertDialog.Builder(this);
    wrongPasswordBuilder.setTitle(getString(R.string.wrongPasswordTitle));
    wrongPasswordBuilder.setMessage(getString(R.string.wrongPasswordTryAgain));
    wrongPasswordBuilder.setPositiveButton("ok", null);
    AlertDialog dialog = wrongPasswordBuilder.show();
    }
}

简单的android密码检查器无法正常工作

3 个答案: