我的活动中有一些碎片
[1], [2], [3], [4], [5], [6]
如果当前活动片段是[2],那么在返回按钮上按下我必须从[2]返回到[1],否则什么也不做。
最好的做法是什么?
编辑:应用程序不能从[3]…[6]返回[2]
我的活动中有一些碎片
[1], [2], [3], [4], [5], [6]
如果当前活动片段是[2],那么在返回按钮上按下我必须从[2]返回到[1],否则什么也不做。
最好的做法是什么?
编辑:应用程序不能从[3]…[6]返回[2]
当前回答
我宁愿这样做:
private final static String TAG_FRAGMENT = "TAG_FRAGMENT";
private void showFragment() {
final Myfragment fragment = new MyFragment();
final FragmentTransaction transaction = getSupportFragmentManager().beginTransaction();
transaction.replace(R.id.fragment, fragment, TAG_FRAGMENT);
transaction.addToBackStack(null);
transaction.commit();
}
@Override
public void onBackPressed() {
final Myfragment fragment = (Myfragment) getSupportFragmentManager().findFragmentByTag(TAG_FRAGMENT);
if (fragment.allowBackPressed()) { // and then you define a method allowBackPressed with the logic to allow back pressed or not
super.onBackPressed();
}
}
其他回答
当你在Fragments之间转换时,调用addToBackStack()作为FragmentTransaction的一部分:
FragmentTransaction tx = fragmentManager.beginTransation();
tx.replace( R.id.fragment, new MyFragment() ).addToBackStack( "tag" ).commit();
如果你需要更详细的控制(例如,当一些片段可见时,你想要抑制返回键),你可以在你的片段的父视图上设置一个OnKeyListener:
//You need to add the following line for this solution to work; thanks skayred
fragment.getView().setFocusableInTouchMode(true);
fragment.getView().requestFocus();
fragment.getView().setOnKeyListener( new OnKeyListener()
{
@Override
public boolean onKey( View v, int keyCode, KeyEvent event )
{
if( keyCode == KeyEvent.KEYCODE_BACK )
{
return true;
}
return false;
}
} );
对于那些谁使用静态片段
在这种情况下,如果你有一个静态片段,那么它会更可取。 为片段创建一个实例对象
private static MyFragment instance=null;
在MyFragment的onCreate()中初始化该实例
instance=this;
也可以创建一个函数来获取Instance
public static MyFragment getInstance(){
return instance;
}
也可以创建函数
public boolean allowBackPressed(){
if(allowBack==true){
return true;
}
return false;
}
//allowBack is a boolean variable that will be set to true at the action
//where you want that your backButton should not close activity. In my case I open
//Navigation Drawer then I set it to true. so when I press backbutton my
//drawer should be get closed
public void performSomeAction(){
//.. Your code
///Here I have closed my drawer
}
在你能做的活动中
@Override
public void onBackPressed() {
if (MyFragment.getInstance().allowBackPressed()) {
MyFragment.getInstance().performSomeAction();
}
else{
super.onBackPressed();
}
}
我宁愿这样做:
private final static String TAG_FRAGMENT = "TAG_FRAGMENT";
private void showFragment() {
final Myfragment fragment = new MyFragment();
final FragmentTransaction transaction = getSupportFragmentManager().beginTransaction();
transaction.replace(R.id.fragment, fragment, TAG_FRAGMENT);
transaction.addToBackStack(null);
transaction.commit();
}
@Override
public void onBackPressed() {
final Myfragment fragment = (Myfragment) getSupportFragmentManager().findFragmentByTag(TAG_FRAGMENT);
if (fragment.allowBackPressed()) { // and then you define a method allowBackPressed with the logic to allow back pressed or not
super.onBackPressed();
}
}
如果你想处理硬件返回键事件,那么你必须在Fragment的onActivityCreated()方法中执行以下代码。
你还需要检查Action_Down或Action_UP事件。如果你不检查,那么onKey()方法将调用2次。
同样,如果你的rootview(getView())将不包含焦点,那么它将无法工作。如果你点击了任何控件,那么你需要再次使用getView().requestFocus()给rootview的焦点;在此之后,只有onKeydown()将调用。
getView().setFocusableInTouchMode(true);
getView().requestFocus();
getView().setOnKeyListener(new OnKeyListener() {
@Override
public boolean onKey(View v, int keyCode, KeyEvent event) {
if (event.getAction() == KeyEvent.ACTION_DOWN) {
if (keyCode == KeyEvent.KEYCODE_BACK) {
Toast.makeText(getActivity(), "Back Pressed", Toast.LENGTH_SHORT).show();
return true;
}
}
return false;
}
});
对我来说很好。
检查后台工作完美
@Override
public boolean onKeyDown(int keyCode, KeyEvent event)
{
if (keyCode == KeyEvent.KEYCODE_BACK)
{
if (getFragmentManager().getBackStackEntryCount() == 1)
{
// DO something here since there is only one fragment left
// Popping a dialog asking to quit the application
return false;
}
}
return super.onKeyDown(keyCode, event);
}