2013-07-16 35 views
2

我添加一个可绘制的图像作为标题在列表视图。在活动的情况下,我已经实施了如下。Android添加头到列表视图片段

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

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

不过,现在我想实现同在这样

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; 
    } 
} 

现在,在这个viewpager片段,我得到一个错误“

的方法getLayoutInflater( Bundle)中的类型片段不是 适用于参数()

“在线

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

如何解决此错误? 感谢

回答

3

变化

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

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

onCreatView第一放慢参数当然是LayoutInflater

+0

肯定。这取决于你是如何编写R.layout.header – Blackbelt

+0

好的。再次感谢。 –