ListView Display and File to Uri

How can adapter display the content of the listview.

1.AbsListView.java

[cpp] view plaincopy

1. protected void onMeasure(int widthMeasureSpec, int heightMeasureSpec) {  
2.   .......  
3.   final View child = obtainView(0, mIsScrap);  
4.   .....  
5. }  

//one item, called one time.
2.AbsListView.java

[java] view plaincopy

1.     View obtainView(int position, boolean[] isScrap) {  
2. ....  
3. child = mAdapter.getView(position, scrapView, this);  
4. ....  
5. }  

3.CursorAdapter.java

[java] view plaincopy

1.     public View getView(int position, View convertView, ViewGroup parent) {  
2.         .....  
3.         if (convertView == null) {  
4.             v = newView(mContext, mCursor, parent);  
5.         } else {  
6.             v = convertView;  
7.         }  
8.         bindView(v, mContext, mCursor);  
9.         return v;  
10.     }  

4.CursorAdapter.java

[java] view plaincopy

1. public abstract void bindView(View view, Context context, Cursor cursor);  

[java] view plaincopy

1. public abstract View newView(Context context, Cursor cursor, ViewGroup parent);  

 

//so override the the bindView and newView. and we just need the return the item view not the listview. 

all adapters getView

[java] view plaincopy

1. View getView(int position, View convertView, ViewGroup parent);  

so for what adapter, we just need to return one item's view.

 

file to url

[java] view plaincopy

1. public static Uri fromFile(File file) {  
2.     if (file == null) {  
3.         throw new NullPointerException("file");  
4.     }  
5.   
6.     PathPart path = PathPart.fromDecoded(file.getAbsolutePath());  
7.     return new HierarchicalUri(  
8.             "file", Part.EMPTY, path, Part.NULL, Part.NULL);  
9. } 
   

来源：https://www.cnblogs.com/elfylin/archive/2012/04/06/2435356.html