Question

该应用程序正在使用FragmentStatePagerAdapter和随附的ViewPager。我有3个片段。我试图在第3个片段ViewPager中调用FragmentActivity类（SignUpActivity）中定义的onCreateView()，并通过以下行将ViewPagers当前项设置为第一页：

SignUpActivity.pager.setCurrentItem(0));

以上行导致以下错误：

java.lang.IllegalStateException: Recursive entry to executePendingTransactions

但是通过实验，当我运行与上面相同的setCurrentItem(0)行时，在setUserVisibleHint()方法中，ViewPager按预期显示了第0个片段。为什么ViewPager在第一种情况下失败，我假设ViewPager在完成之前的任何调用之前都无法调用？

我如何能够创建一种机制来从ViewPager和FragmentActivity调用setCurrentItem(x)到我想要的任何视图？

SignUpActivity代码（FragmentActivity）：

public class SignUpActivity extends FragmentActivity {
CustFSPA adapter;
static ViewPager pager;

@Override
public void onCreate(Bundle savedInstanceState){
    super.onCreate(savedInstanceState);
    setContentView(R.layout.sign_up); /

    adapter = new CustFSPA(getSupportFragmentManager());
    pager = (ViewPager) findViewById(R.id.signUpPager);
    pager.setAdapter(adapter);
}

public static class CustFSPA extends FragmentStatePagerAdapter{
    int numberFragments = 3;
    static FragmentManager fragmentManager;

    public CustFSPA(FragmentManager manager){
        super(manager);
        fragmentManager = manager;
    }

    @Override
    public int getCount(){
        return numberFragments;
    }

    @Override
    public Fragment getItem(int position){
        switch(position){
            case 0:
                // will display SignUpFragmentFirst
                return SignUpFragmentFirst.calledFromPager();
            case 1:
                return SignUpFragmentSecond.calledFromPager();
            default:
                return SignUpFragmentThird.calledFromPager();
        }
    }    
}

第三段片段代码:(与我提供前两个片段＆＃39;代码无关）

public class SignUpFragmentThird extends Fragment {

@Override
public void onCreate(Bundle savedInstanceState){
    super.onCreate(savedInstanceState);
    System.out.println("onCreate third");

}

@Override
public void onStart(){
    super.onStart();
    System.out.println("Starting 3rd");
}

@Override
public void onPause(){
    super.onPause();
    System.out.println("Paused 3rd");
}

@Override
public void onResume(){
    super.onResume();
    System.out.println("Resumed 3rd");
}

@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState){
    System.out.println("onCreateView third");
    ViewPager pager = (ViewPager) getActivity().findViewById(R.id.signUpPager);

    SignUpActivity.pager.setCurrentItem(0); <======= THIS WAS THE FIRST ATTEMPT WHICH FAILED AND RESULTED IN THE ERROR

    View view = inflater.inflate(R.layout.sign_up_fragment_third, container, false);
    return view;
}

@Override
public void setUserVisibleHint(boolean isVisibleToUser) {
    super.setUserVisibleHint(isVisibleToUser);
    if (isVisibleToUser) {
        System.out.println("THIRD is visible to the user");
        SignUpActivity.pager.setCurrentItem(1); <========= SECOND ATTEMPT WHICH WOKRED SUCCESSFULLY
    } else {
        System.out.println("THIRD is not visible to the user");
    }
}

static SignUpFragmentThird calledFromPager(){
    System.out.println("third case def  " + SignUpActivity.pager.getCurrentItem());
    System.out.println("calledFromPager third");
    SignUpFragmentThird fragment = new SignUpFragmentThird();
    return fragment;
}
}

谢谢

Answer 1

您可以通过此调用将viewPager设置为任何所需的视图。基本上，此侦听器会跟踪viewpager的页面。你需要在Parent Activity中实现它。

viewPager.addOnPageChangeListener(new ViewPager.OnPageChangeListener() {
    public void onPageScrollStateChanged(int state) {

    }

    public void onPageScrolled(int position, float positionOffset, int positionOffsetPixels) {

    }

    public void onPageSelected(int position) {
        Log.i("TAG", "position: " + position);
       //Here you can change the position of the viewPager.
       //you have three fragnment so when the position would be positon=2 then you can set viewPager to any view
       switch(position)
         case 2:
          pager.setCurrentItem(0));
         break;
    }
});

希望它会有所帮助。

java.lang.IllegalStateException：executePendingTransactions的递归入口，在setUserVisibleHint中调用时不会发生错误

1 个答案: