Android TabsPagerAdapter类错误“;“类型不匹配”;
我在TabsPagerAdapter类中遇到问题,错误显示…Android TabsPagerAdapter类错误“;“类型不匹配”;,android,layout,android-fragments,Android,Layout,Android Fragments,我在TabsPagerAdapter类中遇到问题,错误显示…类型不匹配:无法从MapActivity转换为Fragment帮助我解决此问题 注 选项卡SpagerAdapter类是 import android.support.v4.app.Fragment; import android.support.v4.app.FragmentActivity; import android.support.v4.app.FragmentManager; import android.support.v
类型不匹配:无法从MapActivity转换为Fragment
帮助我解决此问题
注
选项卡SpagerAdapter
类是
import android.support.v4.app.Fragment;
import android.support.v4.app.FragmentActivity;
import android.support.v4.app.FragmentManager;
import android.support.v4.app.FragmentPagerAdapter;
public class TabsPagerAdapter extends FragmentPagerAdapter {
public TabsPagerAdapter(FragmentManager fm) {
super(fm);
}
@Override
public Fragment getItem(int index) {
switch (index) {
case 0:
// Map fragment activity
return new MapActivity();
case 1:
// Directions fragment activity
return new Directions();
case 2:
// Places fragment activity
return new Places();
}
return null;
}
@Override
public int getCount() {
// get item count - equal to number of tabs
return 3;
}
}
您必须只返回
片段的实例。但是这里您的MapActivity
是FragmentActivity
而不是Fragment
…好的,我想从activity返回任何类型,如果它是Fragment或FragmentActivity,那么代码中会有什么变化?
import android.support.v4.app.Fragment;
import android.support.v4.app.FragmentActivity;
import android.support.v4.app.FragmentManager;
import android.support.v4.app.FragmentPagerAdapter;
public class TabsPagerAdapter extends FragmentPagerAdapter {
public TabsPagerAdapter(FragmentManager fm) {
super(fm);
}
@Override
public Fragment getItem(int index) {
switch (index) {
case 0:
// Map fragment activity
return new MapActivity();
case 1:
// Directions fragment activity
return new Directions();
case 2:
// Places fragment activity
return new Places();
}
return null;
}
@Override
public int getCount() {
// get item count - equal to number of tabs
return 3;
}
}