显示打开文件的路径

Question

用一个openFileDialog来打开文件，之后用textBox显示所打开的文件的路径，请问怎样实现，最好给出代码

Answer 1

if(openFileDialog.ShowDialog()==DialogResult.OK)
{
   textBox1.Text=openFileDialog.FilePath;
}

Answer 2

TextBox.Text = openFileDialg.FileName

Answer 3

      private void btn1_Click_1(object sender, System.EventArgs e)
        {   
            if(DialogResult.OK==openFileDialog1.ShowDialog(this))
            {
                textBox1.Text=openFileDialog1.FileName;
            }
        }

Answer 4

System.Windows.Forms.DialogResult result = this.openFileDialog1.ShowDialog();
if (result == DialogResult.OK || result == DialogResult.Yes)
{
this.textBox1.Text = this.openFileDialog1.FileName;
}