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

在fragment类中,为back event放以下代码:

 rootView.setFocusableInTouchMode(true);
        rootView.requestFocus();
        rootView.setOnKeyListener( new OnKeyListener()
        {
            @Override
            public boolean onKey( View v, int keyCode, KeyEvent event )
            {
                if( keyCode == KeyEvent.KEYCODE_BACK )
                {
                    FragmentManager fragmentManager = getFragmentManager();
                    fragmentManager.beginTransaction()
                            .replace(R.id.frame_container, new Book_service_provider()).commit();

                    return true;
                }
                return false;
            }
        } );

其他回答

最理想的方法如下: 片段:当按下后退按钮并自定义时调用的回调

public class MyActivity extends Activity
{
    //...
    //Defined in Activity class, so override
    @Override
    public void onBackPressed()
    {
        super.onBackPressed();
        myFragment.onBackPressed();
    }
}

public class MyFragment extends Fragment
{
    //Your created method
    public static void onBackPressed()
    {
        //Pop Fragments off backstack and do your other checks
    }
}

工作代码:

package com.example.keralapolice;

import android.app.Fragment;
import android.app.FragmentManager;
import android.app.FragmentManager.OnBackStackChangedListener;
import android.content.Intent;
import android.os.Bundle;
import android.util.Log;
import android.view.Gravity;
import android.view.KeyEvent;
import android.view.LayoutInflater;
import android.view.View;
import android.view.ViewGroup;
import android.widget.Toast;

public class ChiefFragment extends Fragment {
    View view;

    // public OnBackPressedListener onBackPressedListener;

    @Override
    public View onCreateView(LayoutInflater inflater,
            ViewGroup container, Bundle args) {

        view = inflater.inflate(R.layout.activity_chief, container, false);
        getActivity().getActionBar().hide();
        view.setFocusableInTouchMode(true);
        view.requestFocus();
        view.setOnKeyListener(new View.OnKeyListener() {
            @Override
            public boolean onKey(View v, int keyCode, KeyEvent event) {
                Log.i(getTag(), "keyCode: " + keyCode);
                if (keyCode == KeyEvent.KEYCODE_BACK) {
                    getActivity().getActionBar().show();
                    Log.i(getTag(), "onKey Back listener is working!!!");
                    getFragmentManager().popBackStack(null, FragmentManager.POP_BACK_STACK_INCLUSIVE);
                    // String cameback="CameBack";
                    Intent i = new Intent(getActivity(), home.class);
                    // i.putExtra("Comingback", cameback);
                    startActivity(i);
                    return true;
                } else {
                    return false;
                }
            }
        });
        return view;
    }
}

在你的oncreateView()方法中,你需要写这些代码,在KEYCODE_BACk条件下,你可以写任何你想要的功能

View v = inflater.inflate(R.layout.xyz, container, false);
//Back pressed Logic for fragment 
v.setFocusableInTouchMode(true); 
v.requestFocus(); 
v.setOnKeyListener(new View.OnKeyListener() { 
    @Override 
    public boolean onKey(View v, int keyCode, KeyEvent event) {
        if (event.getAction() == KeyEvent.ACTION_DOWN) {
            if (keyCode == KeyEvent.KEYCODE_BACK) {
                getActivity().finish(); 
                Intent intent = new Intent(getActivity(), MainActivity.class);
                startActivity(intent);

                return true; 
            } 
        } 
        return false; 
    } 
}); 

在看了所有的解决方案后,我意识到有一个更简单的解决方案。

在你的活动的onBackPressed()托管你所有的片段,找到你想要防止反压的片段。如果找到了,就返回。那么popBackStack将永远不会发生在这个片段上。

  @Override
public void onBackPressed() {

        Fragment1 fragment1 = (Fragment1) getFragmentManager().findFragmentByTag(“Fragment1”);
        if (fragment1 != null)
            return;

        if (getFragmentManager().getBackStackEntryCount() > 0){
            getFragmentManager().popBackStack();

        }
}

我正在与SlidingMenu和Fragment一起工作,在这里展示我的案例,希望能帮助到别人。

按[后退]键时的逻辑:

When SlidingMenu shows, close it, no more things to do. Or when 2nd(or more) Fragment showing, slide back to previous Fragment, and no more things to do. SlidingMenu not shows, current Fragment is #0, do the original [Back] key does. public class Main extends SherlockFragmentActivity { private SlidingMenu menu=null; Constants.VP=new ViewPager(this); //Some stuff... @Override public void onBackPressed() { if(menu.isMenuShowing()) { menu.showContent(true); //Close SlidingMenu when menu showing return; } else { int page=Constants.VP.getCurrentItem(); if(page>0) { Constants.VP.setCurrentItem(page-1, true); //Show previous fragment until Fragment#0 return; } else {super.onBackPressed();} //If SlidingMenu is not showing and current Fragment is #0, do the original [Back] key does. In my case is exit from APP } } }