openFiledialog仅为名称

Question

有人能告诉我如何在名字前面只显示名字而不是完整的路径吗？我发现这段代码只显示扩展名，或者只显示没有扩展名的文件名。我想要显示的是扩展名文件的名称。

就像这样: example.txt。

代码：

private void button1_Click(object sender, EventArgs e)
    {
        openFileDialog1.Filter = "Binary Files (.BIN; .md6; .md7)|*.BIN; *.md6; *.md7|All Files (*.*)|*.*";


        {
            if (openFileDialog1.ShowDialog() == DialogResult.OK)
            {
                System.IO.StreamReader sr = new
                System.IO.StreamReader(openFileDialog1.FileName);
                sr.Close();
            }
        }
        String filedata = openFileDialog1.FileName;

        openFileDialog1.Title = ("Choose a file");
        openFileDialog1.InitialDirectory = "C:\\Projects\\flashloader2013\\mainapplication\\Bootfiles";

        //textBox1.Text = (System.IO.Path.GetExtension(openFileDialog1.FileName));
        textBox1.Text = (System.IO.Path.Get(openFileDialog1.FileName));
    }

谢谢大家对的帮助

Answer 1

var onlyfilename = Path.GetFileName(openFileDialog1.FileName);

Answer 2

您签出SafeFileName属性了吗？http://msdn.microsoft.com/en-us/library/system.windows.forms.openfiledialog.safefilename.aspx