Viewpager的NullPointerException内传递微调从片段片段分配值[解决] [英] Passing Assigned Value of a Spinner from Fragment to Fragment within Viewpager NullPointerException [Solved]
问题描述
好吧，我让它更简单，根据我的previous后的答案更新它：<一href=\"http://stackoverflow.com/questions/20649497/adding-assigned-values-in-spinner-nullpointerexception\">Adding在微调的NullPointerException 赋值
Ok, I've make it more simple and updated it based on the answers in my previous post: Adding Assigned Values in Spinner NullPointerException
我有一个 MainAcitivty ，它使用的 ViewPager 。我有2个片段在我MainActivity（弗拉加和FragB）
I have a MainAcitivty that uses a ViewPager. I have 2 Fragments in my MainActivity (FragA and FragB)
在我的弗拉加我有一个微调。而在 FragB ，我有一个的TextView 和按钮即可。
In my FragA I have a spinner. While in FragB, I have a TextView and a Button.
现在我是什么要做的是，当我在飞旋选择你好，我的int类型的将有5的值，当我按一下按钮，5将在TextView中显示。
Now what am I trying to do is, when I select "Hello" in my spinner, my int a will have a value of 5. And when I click the Button, 5 will display in the TextView.
下面是我的code：
弗拉加
public class FragA extends Fragment {
Spinner spinner1;
String s1;
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
Bundle savedInstanceState) {
View view = inflater.inflate(R.layout.fraga, container, false);
spinner1 = (Spinner) view.findViewById(R.id.spinner1);
ArrayAdapter<CharSequence> adapter_a = ArrayAdapter.createFromResource(getActivity(), R.array.spinner1,android.R.layout.simple_spinner_item );
spinner1.setAdapter(adapter_a);
s1 = spinner1.getSelectedItem().toString();
return view;
}
public int getInt() {
int a = 0;
if(s1.equals("Hello")) {
a = 5;
}
return a;
}
}
MainActivity
public class MainActivity extends FragmentActivity {
ViewPager viewPager = null;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
viewPager = (ViewPager)findViewById(R.id.pager);
FragmentManager fragmentManager = getSupportFragmentManager();
viewPager.setAdapter(new MyAdapter(fragmentManager));
}
public class MyAdapter extends FragmentStatePagerAdapter {
public MyAdapter (FragmentManager fm) {
super(fm);
}
@Override
public Fragment getItem(int i) {
Fragment fragment = null;
if (i == 0)
{
fragment = new FragA();
}
if (i == 1)
{
fragment = new FragB();
}
return fragment;
}
@Override
public int getCount() {
return 2;
}
}
public String get() {
FragA FragA = new FragA();
return Integer.toString(FragA.getInt());
}
}
FragB
public class FragB extends Fragment{
TextView textView;
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
Bundle savedInstanceState) {
View view = inflater.inflate(R.layout.fragb, container, false);
textView = (TextView) view.findViewById(R.id.textview);
Button button = (Button) view.findViewById(R.id.button);
button.setOnClickListener(Click);
return view;
}
OnClickListener Click = new OnClickListener() {
@Override
public void onClick(View v) {
textView.setText(((MainActivity)getActivity()).get());
}
};
}
顺便说一句，我有过的从弗拉加到FragB的值没有问题的时候，这是我的code：
By the way, I have no problem passing the value of a from FragA to FragB when this is my code:
public int getInt() {
int a = 5;
return a;
}
但是，这并不涉及我的微调而不是我想做的事情。希望有人可以给我一个解决我的问题。一直停留在此的某个时候。先谢谢了。
But, it doesn't involve my spinner and that's not I want to do. Hope someone can give me a solution to my problem. Been stuck on this for sometime. Thanks in advance.
推荐答案
下面是一个简单的方法片段之间的通信。
Here is a simple way to communicate between fragments.
在MainActivity，保持片段的静态实例。
In MainActivity, keep static instances of the fragments.
public static FragA fragmentA;
public static FragB fragmentB;
现在，如果你想从弗拉加访问FragB，写类似的东西：
Now if you want to access FragB from FragA, write something similar:
((MainActivity) getActivity()).fragmentB.setSomething();
和这是一个更好/有道片段之间进行通信：
And Here is a better/proper way to communicate between fragments:
http://developer.android.com/training/basics/fragments/ communicating.html
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