在viewpager中使用backstack和back按钮

时间:2013-08-26 21:25:37

标签: android android-fragments android-viewpager back-stack

我正在使用viewpager在片段之间滑动,并希望后退按钮导航到先前查看的片段而不是结束活动。很抱歉,如果这是this question的副本,但我没有找到答案非常有帮助。显然onBackPressed需要被覆盖,但我不知道如何获取并显示正确的片段。我假设我应该使用fragmentmanager的backstack,但是getSupportFragmentManager().getBackStackEntryCount()总是返回0.我是否需要使用FragmentTransaction.addToBackStack()手动将片段添加到backstack?如果是这样,我会在适配器中添加它?

以下是我的活动的代码,

public class PagerActivity extends FragmentActivity {

ArrayList<Sale> sales;
MyAdapter mAdapter;
ViewPager mPager;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);

    Intent it = getIntent();
    this.sales = (ArrayList<Sale>) it.getExtras().get("sales");
    int position = it.getExtras().getInt("position");

    ActionBar actionBar = getActionBar();
    actionBar.setDisplayHomeAsUpEnabled(true);

    setContentView(R.layout.fragment_pager);
    mAdapter = new MyAdapter(getSupportFragmentManager());
    mPager = (ViewPager) findViewById(R.id.pager);
    mPager.setAdapter(mAdapter);
    mPager.setCurrentItem(position);

}

public class MyAdapter extends FragmentPagerAdapter {
    public MyAdapter(FragmentManager fragmentManager) {
        super(fragmentManager);
    }

    @Override
    public int getCount() {
        return sales.size();
    }

    @Override
    public void destroyItem(ViewGroup container, int position, Object object) {
        super.destroyItem(container, position, object);
    }

    @Override
    public Fragment getItem(int position) {
        SalesThumbFragment frag = new SalesThumbFragment();
        return frag.newInstance(sales.get(position));
    }
}

@Override
public boolean onOptionsItemSelected(MenuItem item) {
    switch (item.getItemId()) {
    case android.R.id.home:
        finish();
        return true;
    }
    return super.onOptionsItemSelected(item);
}

@Override
public boolean onCreateOptionsMenu(Menu menu) {
    // Inflate the menu; this adds items to the action bar if it is present.
    getMenuInflater().inflate(R.menu.sales_controller, menu);
    return true;
}

@Override
public void onBackPressed() {
  if(getSupportFragmentManager().getBackStackEntryCount() != 0) {
    getSupportFragmentManager().popBackStack();
  } else {

    super.onBackPressed();
  }
}


}

5 个答案:

答案 0 :(得分:8)

在新设计支持库中,我使用此
我有同样的问题,我按照这一步

在viewpager中有3个片段的主要活动中,我创建堆栈 和推送和弹出数据。

//private Stack<Integer> stackkk;  ==> As i get edit suggestion 
private Stack<Integer> stackkk = new Stack<>(); // Edited 
private ViewPager mPager;
private int tabPosition = 0;



    mTabLayout.setupWithViewPager(mPager);
    mPager.addOnPageChangeListener(new TabLayout.TabLayoutOnPageChangeListener(mTabLayout));
    mTabLayout.setOnTabSelectedListener(new TabLayout.OnTabSelectedListener() {
        @Override
        public void onTabSelected(TabLayout.Tab tab) {
            tabPosition = tab.getPosition();
            mPager.setCurrentItem(tab.getPosition());

            if (stackkk.empty())
                stackkk.push(0);

            if (stackkk.contains(tabPosition)) {
                stackkk.remove(stackkk.indexOf(tabPosition));
                stackkk.push(tabPosition);
            } else {
                stackkk.push(tabPosition);
            }
        }

        @Override
        public void onTabUnselected(TabLayout.Tab tab) {
            tabPositionUnselected = tab.getPosition();
        }

        @Override
        public void onTabReselected(TabLayout.Tab tab) {
        }
    });
}

并且在onBackPressed in activity中,

@Override
public void onBackPressed() {
    if (stackkk.size() > 1) {
        stackkk.pop();
        mPager.setCurrentItem(stackkk.lastElement());
    } else {
    }
}

答案 1 :(得分:2)

在Fragment Activity类中使用此代码。不要忘记添加return true;

public boolean onKeyDown(int keyCode, KeyEvent event) {

    // TODO Auto-generated method stub
    if ((keyCode == KeyEvent.KEYCODE_BACK)) {

            mViewPager.setCurrentItem(viewPageSelected - 1);
            return true;
        }

    return super.onKeyDown(keyCode, event);

}

答案 2 :(得分:1)

我有类似的问题,这就是我解决它的方法。如果我明白你的问题,我认为你可以根据你的问题调整代码。我有一个包含6个片段的ViewPager,想要跟踪页面历史记录,并能够使用后退按钮在历史记录中向后导航。我创建一个java.util.Stack<Integer>对象,向其添加片段编号(除非使用后退按钮,见下文),并覆盖onBackPressed()以使其弹出最后查看的片段而不是使用后台堆栈,当我的历史堆栈不为空时。

当您按下后退按钮时,您希望避免在堆栈上推送元素,否则如果继续使用后退按钮,则会卡在两个片段之间,而不是最终退出。

我的代码:

MyAdapter mAdapter;
ViewPager mPager;
Stack<Integer> pageHistory;
int currentPage;
boolean saveToHistory;

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

    mAdapter = new MyAdapter(getSupportFragmentManager());
    mPager = (ViewPager)findViewById(R.id.container);
    mPager.setAdapter(mAdapter);
    mPager.setOffscreenPageLimit(5);

    pageHistory = new Stack<Integer>();
    mPager.setOnPageChangeListener(new OnPageChangeListener() {

        @Override
        public void onPageSelected(int arg0) {
            if(saveToHistory)
                pageHistory.push(Integer.valueOf(currentPage));
        }

        @Override
        public void onPageScrolled(int arg0, float arg1, int arg2) {
        }

        @Override
        public void onPageScrollStateChanged(int arg0) {
        }
    });
    saveToHistory = true;
}

@Override
public void onBackPressed() {
    if(pageHistory.empty())
        super.onBackPressed();
    else {
        saveToHistory = false;
        mPager.setCurrentItem(pageHistory.pop().intValue());
        saveToHistory = true;
    }
};

答案 3 :(得分:0)

如果您在每次用户导航到新片段后使用字段跟踪上一页的索引,则使用mPager.getCurrentItem(),然后在onBackPressed()方法中,您应该能够致电mPager.setCurrentItem(previousPage)

或者,如果用户只能按顺序分页,那么根本不需要字段,您可以mPager.setCurrentItem(mPager.getCurrentItem()-1)

答案 4 :(得分:0)

我已经制作了自定义ViewPager并在其中实现了堆栈功能。

public class CustomViewPager extends ViewPager {

private Stack<Integer> stack = new Stack<>();

@Override
public void setCurrentItem(int item, boolean smoothScroll) {
    stack.push(getCurrentItem());
    super.setCurrentItem(item, smoothScroll);
}

public int popFromBackStack(boolean smoothScroll) {
    if (stack.size()>0) {
        super.setCurrentItem(stack.pop(), smoothScroll);
        return getCurrentItem();
    } else return -1;
}