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我總共有超過1000個標籤/頁面可供查看。我希望能夠通過向EditText輸入一個數字來查看任何標籤/頁面,並且當點擊按鈕時,它應該跳轉到指定的標籤。例如,如果我在第1頁上,並且如果我輸入「600」並點擊搜索(按鈕),我應該能夠查看該特定頁面並確保600旁邊的頁面應該是601,601 ....以及之前600應該是599,598 ....使用按鈕切換標籤通過輸入標籤號碼轉到另一個標籤?
public class MainActivity extends FragmentActivity {
SectionsPagerAdapter mSectionsPagerAdapter;
ViewPager mViewPager;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
mSectionsPagerAdapter = new SectionsPagerAdapter(getSupportFragmentManager());
mViewPager = (ViewPager) findViewById(R.id.pager);
mViewPager.setAdapter(mSectionsPagerAdapter);
}
@Override
public boolean onCreateOptionsMenu(Menu menu) {
getMenuInflater().inflate(R.menu.main, menu);
return true;
}
public class SectionsPagerAdapter extends FragmentPagerAdapter {
public SectionsPagerAdapter(FragmentManager fm) {
super(fm);
}
public Fragment getItem(int position) {
switch(position){
case 0:
case 1:
Ang100s.Ang1 ang1 = new Ang100s.Ang1();
return ang1;
case 2:
Ang100s.Ang2 ang2 = new Ang100s.Ang2();
return ang2;
case 3:
Ang100s.Ang3 ang4 = new Ang100s.Ang3();
return ang4;
}
Help help = new Help();
return help;
}
@Override
public int getCount() {
return 1430;
}
@Override
public CharSequence getPageTitle(int position) {
Locale l = Locale.getDefault();
for(int i = 1;i < 1430; i++){
if(position == i){
return "Ang " + position;
}
}
return null;
}
}
public static class Help extends Fragment {
public View onCreateView(LayoutInflater inflater, ViewGroup container,
Bundle savedInstanceState) {
View rootView = inflater.inflate(R.layout.fragment_main_dummy,
container, false);
final Button search = (Button) rootView.findViewById(R.id.clicktosearch);
final EditText number = (EditText)rootView.findViewById(R.id.number);
search.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
// TODO Auto-generated method stub
if(number.getText().toString().equals("1")){
}
}
});
return rootView;
}
}
}
我試過在OnClick(View v)方法下的Help類中通過實例化 MainActivity main = new MainActivity(); main.mViewPager.setCurrentItem(1); – ivesingh
這不起作用。你應該只有一個MainActivity,你需要調用它的ViewPager。創建另一個MainActivity實例不會做任何有價值的事情(很可能會遇到錯誤)。只有Android操作系統應該實例化你的活動。 – Karakuri
明白了。非常感謝你的幫助 – ivesingh