我试图使用片段与一个ViewPager使用FragmentPagerAdapter。
我想要实现的是替换一个片段，位于ViewPager的第一页，与另一个。
寻呼机由两个页面组成。第一个是FirstPagerFragment，第二个是SecondPagerFragment。点击第一页的一个按钮。我想用NextFragment替换FirstPagerFragment。
下面是我的代码。
public class FragmentPagerActivity extends FragmentActivity {
static final int NUM_ITEMS = 2;
MyAdapter mAdapter;
ViewPager mPager;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.fragment_pager);
mAdapter = new MyAdapter(getSupportFragmentManager());
mPager = (ViewPager) findViewById(R.id.pager);
mPager.setAdapter(mAdapter);
}
/**
* Pager Adapter
*/
public static class MyAdapter extends FragmentPagerAdapter {
public MyAdapter(FragmentManager fm) {
super(fm);
}
@Override
public int getCount() {
return NUM_ITEMS;
}
@Override
public Fragment getItem(int position) {
if(position == 0) {
return FirstPageFragment.newInstance();
} else {
return SecondPageFragment.newInstance();
}
}
}
/**
* Second Page FRAGMENT
*/
public static class SecondPageFragment extends Fragment {
public static SecondPageFragment newInstance() {
SecondPageFragment f = new SecondPageFragment();
return f;
}
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
//Log.d("DEBUG", "onCreateView");
return inflater.inflate(R.layout.second, container, false);
}
}
/**
* FIRST PAGE FRAGMENT
*/
public static class FirstPageFragment extends Fragment {
Button button;
public static FirstPageFragment newInstance() {
FirstPageFragment f = new FirstPageFragment();
return f;
}
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
//Log.d("DEBUG", "onCreateView");
View root = inflater.inflate(R.layout.first, container, false);
button = (Button) root.findViewById(R.id.button);
button.setOnClickListener(new OnClickListener() {
@Override
public void onClick(View v) {
FragmentTransaction trans = getFragmentManager().beginTransaction();
trans.replace(R.id.first_fragment_root_id, NextFragment.newInstance());
trans.setTransition(FragmentTransaction.TRANSIT_FRAGMENT_OPEN);
trans.addToBackStack(null);
trans.commit();
}
});
return root;
}
/**
* Next Page FRAGMENT in the First Page
*/
public static class NextFragment extends Fragment {
public static NextFragment newInstance() {
NextFragment f = new NextFragment();
return f;
}
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
//Log.d("DEBUG", "onCreateView");
return inflater.inflate(R.layout.next, container, false);
}
}
}
.．.这里是XML文件
fragment_pager.xml
<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:orientation="vertical" android:padding="4dip"
android:gravity="center_horizontal"
android:layout_width="match_parent" android:layout_height="match_parent">
<android.support.v4.view.ViewPager
android:id="@+id/pager"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:layout_weight="1">
</android.support.v4.view.ViewPager>
</LinearLayout>
first.xml
<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:id="@+id/first_fragment_root_id"
android:orientation="vertical"
android:layout_width="match_parent"
android:layout_height="match_parent">
<Button android:id="@+id/button"
android:layout_width="wrap_content" android:layout_height="wrap_content"
android:text="to next"/>
</LinearLayout>
现在的问题是……我应该使用哪个ID
trans.replace(R.id.first_fragment_root_id, NextFragment.newInstance());
?
如果我用r。id。first_fragment_root_id，替换工作，但Hierarchy Viewer显示一个奇怪的行为，如下所示。
一开始的情况是
更换后的情况是
正如你所看到的，有一些错误，我希望在替换片段后找到与第一张图片中相同的状态。
Works Great with AndroidTeam's solution, however I found that I needed the ability to go back much like FrgmentTransaction.addToBackStack(null) But merely adding this will only cause the Fragment to be replaced without notifying the ViewPager. Combining the provided solution with this minor enhancement will allow you to return to the previous state by merely overriding the activity's onBackPressed() method. The biggest drawback is that it will only go back one at a time which may result in multiple back clicks
private ArrayList<Fragment> bFragments = new ArrayList<Fragment>();
private ArrayList<Integer> bPosition = new ArrayList<Integer>();
public void replaceFragmentsWithBackOut(ViewPager container, Fragment oldFragment, Fragment newFragment) {
startUpdate(container);
// remove old fragment
if (mCurTransaction == null) {
mCurTransaction = mFragmentManager.beginTransaction();
}
int position = getFragmentPosition(oldFragment);
while (mSavedState.size() <= position) {
mSavedState.add(null);
}
//Add Fragment to Back List
bFragments.add(oldFragment);
//Add Pager Position to Back List
bPosition.add(position);
mSavedState.set(position, null);
mFragments.set(position, null);
mCurTransaction.remove(oldFragment);
// add new fragment
while (mFragments.size() <= position) {
mFragments.add(null);
}
mFragments.set(position, newFragment);
mCurTransaction.add(container.getId(), newFragment);
finishUpdate(container);
// ensure getItem returns newFragemtn after calling handleGetItemInbalidated()
handleGetItemInvalidated(container, oldFragment, newFragment);
container.notifyItemChanged(oldFragment, newFragment);
}
public boolean popBackImmediate(ViewPager container){
int bFragSize = bFragments.size();
int bPosSize = bPosition.size();
if(bFragSize>0 && bPosSize>0){
if(bFragSize==bPosSize){
int last = bFragSize-1;
int position = bPosition.get(last);
//Returns Fragment Currently at this position
Fragment replacedFragment = mFragments.get(position);
Fragment originalFragment = bFragments.get(last);
this.replaceFragments(container, replacedFragment, originalFragment);
bPosition.remove(last);
bFragments.remove(last);
return true;
}
}
return false;
}
希望这能帮助到一些人。
同样，就getFragmentPosition()而言，它几乎是反向的getItem()。你知道哪些片段去哪里，只是确保你返回正确的位置它将在。这里有一个例子:
@Override
protected int getFragmentPosition(Fragment fragment) {
if(fragment.equals(originalFragment1)){
return 0;
}
if(fragment.equals(replacementFragment1)){
return 0;
}
if(fragment.equals(Fragment2)){
return 1;
}
return -1;
}
要替换ViewPager中的片段，你可以将ViewPager、PagerAdapter和FragmentStatePagerAdapter类的源代码移动到你的项目中，并添加以下代码。
分为:viewpage:
public void notifyItemChanged(Object oldItem, Object newItem) {
if (mItems != null) {
for (ItemInfo itemInfo : mItems) {
if (itemInfo.object.equals(oldItem)) {
itemInfo.object = newItem;
}
}
}
invalidate();
}
FragmentStatePagerAdapter:
public void replaceFragmetns(ViewPager container, Fragment oldFragment, Fragment newFragment) {
startUpdate(container);
// remove old fragment
if (mCurTransaction == null) {
mCurTransaction = mFragmentManager.beginTransaction();
}
int position = getFragmentPosition(oldFragment);
while (mSavedState.size() <= position) {
mSavedState.add(null);
}
mSavedState.set(position, null);
mFragments.set(position, null);
mCurTransaction.remove(oldFragment);
// add new fragment
while (mFragments.size() <= position) {
mFragments.add(null);
}
mFragments.set(position, newFragment);
mCurTransaction.add(container.getId(), newFragment);
finishUpdate(container);
// ensure getItem returns newFragemtn after calling handleGetItemInbalidated()
handleGetItemInbalidated(container, oldFragment, newFragment);
container.notifyItemChanged(oldFragment, newFragment);
}
protected abstract void handleGetItemInbalidated(View container, Fragment oldFragment, Fragment newFragment);
protected abstract int getFragmentPosition(Fragment fragment);
handleGetItemInvalidated()确保下一次调用getItem()后返回newFragment
getFragmentPosition()返回片段在适配器中的位置。
现在，替换片段调用
mAdapter.replaceFragmetns(mViewPager, oldFragment, newFragment);
如果你对一个示例项目感兴趣，请向我索取资源。
Works Great with AndroidTeam's solution, however I found that I needed the ability to go back much like FrgmentTransaction.addToBackStack(null) But merely adding this will only cause the Fragment to be replaced without notifying the ViewPager. Combining the provided solution with this minor enhancement will allow you to return to the previous state by merely overriding the activity's onBackPressed() method. The biggest drawback is that it will only go back one at a time which may result in multiple back clicks
private ArrayList<Fragment> bFragments = new ArrayList<Fragment>();
private ArrayList<Integer> bPosition = new ArrayList<Integer>();
public void replaceFragmentsWithBackOut(ViewPager container, Fragment oldFragment, Fragment newFragment) {
startUpdate(container);
// remove old fragment
if (mCurTransaction == null) {
mCurTransaction = mFragmentManager.beginTransaction();
}
int position = getFragmentPosition(oldFragment);
while (mSavedState.size() <= position) {
mSavedState.add(null);
}
//Add Fragment to Back List
bFragments.add(oldFragment);
//Add Pager Position to Back List
bPosition.add(position);
mSavedState.set(position, null);
mFragments.set(position, null);
mCurTransaction.remove(oldFragment);
// add new fragment
while (mFragments.size() <= position) {
mFragments.add(null);
}
mFragments.set(position, newFragment);
mCurTransaction.add(container.getId(), newFragment);
finishUpdate(container);
// ensure getItem returns newFragemtn after calling handleGetItemInbalidated()
handleGetItemInvalidated(container, oldFragment, newFragment);
container.notifyItemChanged(oldFragment, newFragment);
}
public boolean popBackImmediate(ViewPager container){
int bFragSize = bFragments.size();
int bPosSize = bPosition.size();
if(bFragSize>0 && bPosSize>0){
if(bFragSize==bPosSize){
int last = bFragSize-1;
int position = bPosition.get(last);
//Returns Fragment Currently at this position
Fragment replacedFragment = mFragments.get(position);
Fragment originalFragment = bFragments.get(last);
this.replaceFragments(container, replacedFragment, originalFragment);
bPosition.remove(last);
bFragments.remove(last);
return true;
}
}
return false;
}
希望这能帮助到一些人。
同样，就getFragmentPosition()而言，它几乎是反向的getItem()。你知道哪些片段去哪里，只是确保你返回正确的位置它将在。这里有一个例子:
@Override
protected int getFragmentPosition(Fragment fragment) {
if(fragment.equals(originalFragment1)){
return 0;
}
if(fragment.equals(replacementFragment1)){
return 0;
}
if(fragment.equals(Fragment2)){
return 1;
}
return -1;
}
