从OpenfileDialog获取文件名称而不是路径

<完整读取文件程序>

private void button1_Click(object sender, EventArgs e)
       {
           OpenFileDialog oFD = new OpenFileDialog();
           oFD.InitialDirectory = Application.StartupPath;
           oFD.Title = "打开文件";
           oFD.Multiselect = true;
           oFD.Filter = "表格文件(*.xls;*.xlsx)|*.xls;*.xlsx|文本文件(*.txt)|*.txt|XML文件(*.xml)|*.xml|所有文件|*.*";
           oFD.FilterIndex = 2;
           oFD.RestoreDirectory = true;
           if(oFD.ShowDialog()==DialogResult.OK)
           {
               string filePath = oFD.FileName;
               string fileName = oFD.SafeFileName;
           }
       }

实例：Form form=new Form();
//OpenFileDialog构造函数
OpenFileDialog dialog = new OpenFileDialog
{
Multiselect = true,//该值确定是否可以选择多个文件
Title = “请选择文件夹”,
Filter = “文件(.xml,.txt)|.xml;.txt”
};
方式1：form.Text=”打开文件：” + dialog.SafeFileName;
方式2：form.Text=”打开文件：” + System.IO.Path.GetFileName(dialog.FileName);
方式3： string s = dialog.FileName;
s = s.Substring(s.LastIndexOf(‘\‘) + 1, s.LastIndexOf(‘.’) - s.LastIndexOf(‘\‘) - 1);
form.Text = “打开文件：” + s;

PS：
OpenFileDialog
OpenFileDialog.SafeFileName 获取对话框中所选文件的文件名和扩展名字，文件名不包含路径
string string.Substring(int StartIndex,int length)从指定字符位置开始，长度为length，得到字符串类型
int string.LastIndexOf（char value）