C# 控件属性编程
请问下,c#在控件属性编程的时候,怎么在属性栏实现OpenFileDialog,效果是,能在属性栏中获得用户打开的文件路径!
OpenFileDialog.FileName

程序代码: [DisplayName("文章内容")]
[Description("文章内容,String类型")]
[TypeConverter(typeof(TextConverter))]
[Editor(typeof(TextTypeEditor), typeof(UITypeEditor))]
public string Context
{
get
{
return _context;
}
set
{
_context = value;
}
}
