Android PagerAdapter 起始位置
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PagerAdapter start position
提问by Kirill Kulakov
I'm using the following example to impliment my viewPager: http://code.google.com/p/viewpagerexample/issues/list
我正在使用以下示例来实现我的 viewPager：http: //code.google.com/p/viewpagerexample/issues/list
The problem with this example is that I can't figure out how to set my starting position, the default starting position is 0. Basically I wan't to be able to control if there is an available view on its left or the right.
这个例子的问题是我不知道如何设置我的起始位置，默认的起始位置是 0。基本上我无法控制它的左侧或右侧是否有可用的视图。
Is there any way to control center's View current position? is there a better way to do it? is it possible to make it circular?
有没有办法控制中心的查看当前位置？有没有更好的方法来做到这一点？有没有可能让它变成圆形？
回答by Kirill Kulakov
I've found a way to set it's position, which is done outside of the class:
我找到了一种方法来设置它的位置，这是在课外完成的：
awesomePager = (ViewPager) findViewById(R.id.awesomepager);
awesomePager.setAdapter(awesomeAdapter);
awesomePager.setCurrentItem(CurrentPosition);
and it can be limited by calculating the amount of items I want to fit in to it
并且可以通过计算我想要放入的项目数量来限制它
回答by Malachiasz
I have noticed that if you recreate Activity (orientation change) with ViewPager having FragmentStatePagerAdapter, then the Adapter will reuse it's Fragments. The way to stop it is:
我注意到，如果您使用具有 FragmentStatePagerAdapter 的 ViewPager 重新创建 Activity（方向更改），则适配器将重用它的 Fragments。阻止它的方法是：
@Override
public void onSaveInstanceState(Bundle savedInstanceState) {
if (viewPager != null) {
// before screen rotation it's better to detach pagerAdapter from the ViewPager, so
// pagerAdapter can remove all old fragments, so they're not reused after rotation.
viewPager.setAdapter(null);
}
super.onSaveInstanceState(savedInstanceState);
}
but then after Activity recreation ViewPager alwayes opens page 0 first and setCurrentItem(CurrentPosition); doesn't work. Fix for that is changing page after delay:
但是在 Activity 娱乐 ViewPager 之后总是先打开第 0 页并 setCurrentItem(CurrentPosition); 不起作用。修复延迟后更改页面的问题：
new Handler().postDelayed(new Runnable() {
@Override
public void run() {
viewPager.setCurrentItem(newPosition);
}
}, 100);
回答by Helali
To start with the last fragment I did this:
从最后一个片段开始，我是这样做的：
PagerAdapter pagerAdapter = new PagerAdapter();
viewPager.setAdapter(pagerAdapter);
viewPager.setCurrentItem(pagerAdapter.getCount() - 1);
回答by DaveB
I came across a problem whereby if I set the current item before I set the adapter, the first item I get back will always be the one at position 0.
我遇到了一个问题，如果我在设置适配器之前设置当前项目，我返回的第一个项目将始终是位置 0 的项目。
Make sure you do:
确保你这样做：
awesomePager.setAdapter(awesomeAdapter);
awesomePager.setCurrentItem(CurrentPosition);
and not:
并不是：
awesomePager.setCurrentItem(CurrentPosition);
awesomePager.setAdapter(awesomeAdapter);
回答by Suhail k khan
I have an array of size more than 1000 and in dymanic viewpager I was facing leftswipe stuck on firstload. The below code solved this and resulted in smooth scroll:
我有一个大小超过 1000 的数组，在 dymanic viewpager 中，我面临着左滑动卡在第一次加载时的问题。下面的代码解决了这个问题并导致平滑滚动：
@Override
onResume(){
super.onResume();
viewPager.setCurrentItem(newPosition);
}
回答by Leebeedev
I'm using 3 fragments and on starting my app, the second (middle) fragment will be shown by default. Just I'm using the onResume function and all works great.
我正在使用 3 个片段，在启动我的应用程序时，默认情况下将显示第二个（中间）片段。只是我正在使用 onResume 函数，一切都很好。
@Override
protected void onResume() {
super.onResume();
m_viewPager.setCurrentItem(1);
}
回答by chanzmao
I encountered same problem. When I initialize ViewPager, the indicator position was 0. This may depends on amount of calculating for Pager contents. I use ViewTreeObserver like below.
我遇到了同样的问题。当我初始化 ViewPager 时，指标位置为 0。这可能取决于 Pager 内容的计算量。我使用如下所示的 ViewTreeObserver。
mGlobalLayoutListener = new ViewTreeObserver.OnGlobalLayoutListener() {
@Override
public void onGlobalLayout() {
mViewPager.setCurrentItem(newPosition);
removeOnGlobalLayoutListener(mSlidingTabLayout.getViewTreeObserver(), mGlobalLayoutListener);
}
};
mSlidingLayout.getViewTreeObserver().addOnGlobalLayoutListener(mGlobalLayoutListener);
and,
和，
private void removeOnGlobalLayoutListener(ViewTreeObserver observer, ViewTreeObserver.OnGlobalLayoutListener listener) {
if (observer == null) return;
if (Build.VERSION.SDK_INT < Build.VERSION_CODES.JELLY_BEAN) {
observer.removeGlobalOnLayoutListener(listener);
} else {
observer.removeOnGlobalLayoutListener(listener);
}
}
In this way, also never to bother with time setting of delay.
这样，也不必费心延迟时间设置。
回答by Yousef Gamal
Using viewPager.setCurrentItem(newPosition);shows to the user the transition from the starting page to the newPosition, to prevent that from happening and show the newPositiondirectly as if it was the starting point, I added falseto the second parameter something like this:
使用viewPager.setCurrentItem(newPosition);向用户显示从起始页到 的转换newPosition，为了防止这种情况发生并将newPosition直接显示为起点，我false在第二个参数中添加了如下内容：
int newPosition = pages.size()-1; // Get last page position
viewPager.setCurrentItem(newPosition, false); // 2nd parameter (false) stands for "smoothScroll"
回答by Kvant
@Override
public Object instantiateItem(ViewGroup container, int position) {
View currentPage = null;
switch(position){
case 0:
currentPage = LayoutInflater.from(context).inflate(R.layout.page0, null)
break;
case 1:
currentPage = LayoutInflater.from(context).inflate(R.layout.page1, null)
///////////// This page will be default ////////////////////
((ViewPager)container).setCurrentItem(position);
////////////////////////////////////////////////////////////
break;
case 2:
currentPage = LayoutInflater.from(context).inflate(R.layout.page2, null)
break;
return currentPage;
}
