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我遇到一个点击监听器的登录模式提交按钮的问题.
这是错误.
Caused by: java.lang.NullPointerException: Attempt to invoke virtual method 'void android.widget.Button.setOnClickListener(android.view.View$OnClickListener)' on a null object reference
我有一个合理的理解,一个空指针异常是什么,我有一个类似于我的问题彻底搜索.我已经尝试以多种方式重新格式化点击监听器,确保我有正确的视图ID等.
package...
import...
public class MainActivity extends ActionBarActivity implements NavigationDrawerFragment.NavigationDrawerCallbacks {
//Variables
String currentPage = "";
Stack crumbs = new Stack();
//Fragment managing the behaviors,interactions and presentation of the navigation drawer.
private NavigationDrawerFragment mNavigationDrawerFragment;
// Used to store the last screen title. For use in {@link #restoreActionBar()}.
public CharSequence mTitle;
//temp
AuthenticateUserTokenResult authenticateUserTokenResult;
String loginErrorMessage = "";
String loginErrorTitle = "";
Boolean logonSuccessful = false;
Dialog loginDialog;
// Login EditTexts
EditText Username;
EditText CompanyID;
EditText Password;
Button Submit;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
mNavigationDrawerFragment = (NavigationDrawerFragment) getSupportFragmentManager().findFragmentById(R.id.navigation_drawer);
mTitle = getTitle(); // Set up the drawer.
mNavigationDrawerFragment.setUp(R.id.navigation_drawer,(DrawerLayout) findViewById(R.id.drawer_layout));
if(authenticateUserTokenResult == null) {
attemptLogin();
}
}
public void attemptLogin() {
loginDialog = new Dialog(this,android.R.style.Theme_Translucent_NoTitleBar);
loginDialog.requestWindowFeature(Window.FEATURE_NO_TITLE);
loginDialog.setContentView(R.layout.login_modal);
loginDialog.setCancelable(false);
//loginDialog.setOnCancelListener(cancelListener);
loginDialog.show();
Submit = (Button)findViewById(R.id.Submit);
Submit.setOnClickListener(new View.OnClickListener() // the error is on this line (specifically the .setOnClickListener)
{
@Override
public void onClick(View v)
{
ClyxUserLogin user = new ClyxUserLogin();
Username = (EditText)findViewById(R.id.Username);
user.logon = Username.getText().toString();
CompanyID = (EditText)findViewById(R.id.CompanyID);
user.idCompany = Integer.parseInt(CompanyID.getText().toString());
Password = (EditText)findViewById(R.id.Password);
user.password = Password.getText().toString();
user.idApplication = 142;
authenticate(user);
}
});
}
显而易见,但与我认为的话题无关.
以下是对话框的XML文件,其上有按钮.
android:orientation="vertical"
android:layout_width="fill_parent"
android:layout_height="fill_parent"
android:background="#3366FF">
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:layout_centerInParent="true"
android:background="#FFFFFF" >
android:id="@+id/LoginTitle"
android:layout_width="200dp"
android:layout_height="wrap_content"
android:gravity="center_horizontal"
android:layout_marginTop="10dp"
android:layout_marginStart="10dp"
android:textColor="#000000"
android:textSize="20sp"
android:text="Login" />
android:id="@+id/Username"
android:layout_width="200dp"
android:layout_height="wrap_content"
android:layout_below="@+id/LoginTitle"
android:layout_margin="10dp"
android:hint="Username" />
android:id="@+id/CompanyID"
android:layout_width="200dp"
android:layout_height="wrap_content"
android:layout_below="@+id/Username"
android:layout_alignStart="@+id/Username"
android:inputType="number"
android:hint="Company ID" />
android:id="@+id/Password"
android:layout_width="200dp"
android:layout_height="wrap_content"
android:layout_below="@+id/CompanyID"
android:layout_alignStart="@+id/Username"
android:layout_marginTop="10dp"
android:layout_marginBottom="10dp"
android:inputType="textPassword"
android:hint="Password" />
android:id="@+id/Submit"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:layout_below="@+id/Password"
android:layout_marginBottom="10dp"
android:layout_centerHorizontal="true"
android:text="Login" />
任何帮助将不胜感激.
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