我在ListView中添加一个可绘制的图像作为标题。我已经实现了如下的情况下的活动。

listView1 = (ListView)findViewById(R.id.listView1);

    View header = (View)getLayoutInflater().inflate(R.layout.header, null);
    listView1.addHeaderView(header);

但是,现在我正试图在这样的viewpager片段中实现相同的功能
import android.os.Bundle;
import android.support.v4.app.Fragment;
import android.view.LayoutInflater;
import android.view.View;
import android.view.ViewGroup;
import android.widget.ArrayAdapter;
import android.widget.ListView;

public class ShowFrag1 extends Fragment {
    @Override
    public View onCreateView(LayoutInflater inflater, ViewGroup container,
        Bundle savedInstanceState) {
        // Inflate the layout for this fragment
        View v=inflater.inflate(R.layout.frag1, container,false);
        ListView lv1=(ListView) v.findViewById(R.id.listView1);

        Level weather_data[] = new Level[]
                {
                    new Level(R.drawable.s1, "L1", R.drawable.p),
                    new Level(R.drawable.s2, "L2",R.drawable.p),
                    new Level(R.drawable.s3, "L3",R.drawable.p),
                    new Level(R.drawable.s4, "L4",R.drawable.p),
                    new Level(R.drawable.s6, "L5",R.drawable.p)
                };
        LevelAdapter adapter = new LevelAdapter(getActivity(),
                R.layout.list_item, weather_data);

       View header = (View)getLayoutInflater().inflate(R.layout.header, null);
        lv1.addHeaderView(header);

        lv1.setAdapter(adapter);
        return v;
    }
}

在这件事上,我犯了个错误”
类型片段中的getLayoutFlater(bundle)方法不是
适用于参数()
“排队
View header = (View)getLayoutInflater().inflate(R.layout.header, null);

如何解决此错误?
谢谢

最佳答案

改变

View header = (View)getLayoutInflater().inflate(R.layout.header, null);

具有
  View header = inflater.inflate(R.layout.header, null);

onCreatView的第一个参数是aLayoutInflater

08-18 17:22