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我有一個查看尋呼機應用程序正常工作。視圖尋呼機包含一個數組中不同的數據串,每次顯示一個數組。我有點擊應該得到的當前內容,並把它傳遞給這裏另一個活動時,一個按鈕是我的代碼:將數據從視圖尋呼機傳遞到不同的活動
/*MainActivity*/
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
// Get the view from viewpager_main.xml
setContentView(R.layout.activity_main);
btn = (Button) findViewById (R.id.button1);
item = new String[] { "Apple", "Movie", "Melon", "pawpaw",
"Jungle", "Forest" };
// Locate the ViewPager in viewpager_main.xml
viewPager = (ViewPager) findViewById(R.id.pager);
// Pass results to ViewPagerAdapter Class
adapter = new ViewPagerAdapter(this, item);
// Binds the Adapter to the ViewPager
viewPager.setAdapter(adapter);
btn.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View arg0) {
// TODO Auto-generated method stub
Intent i = new Intent(getApplicationContext(), ContentActivity.class);
i.putExtra("Currentitem", item[0].toString());
startActivity(i);
}
});
}
下面是ViewPagerAdapter類提前
public class ViewPagerAdapter extends PagerAdapter {
// Declare Variables
Context context;
String[] item;
LayoutInflater inflater;
public ViewPagerAdapter(Context context, String[] item) {
this.context = context;
this.quote = item;
}
@Override
public boolean isViewFromObject(View view, Object object) {
return view == ((RelativeLayout) object);
}
@Override
public Object instantiateItem(ViewGroup container, int position) {
TextView item;
inflater = (LayoutInflater) context
.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
View itemView = inflater.inflate(R.layout.activity_view_pager_adapter, container,
false);
txtitem = (TextView) itemView.findViewById(R.id.quote);
txtitem.setText(quote[position]);
((ViewPager) container).addView(itemView);
return itemView;
感謝
'item [0] .toString()'將總是傳遞數組中的第一項(在這種情況下爲「Apple」)。 – Rajesh
到目前爲止,這是我的,我需要一個代碼,將通過在當前視圖的屏幕上的項目不僅「蘋果」 – Richmahnn
所以你能夠在下一個活動中獲得? –