title: OpenFileDialog Control - How Can I Grab the Selected Path and Show it in a Text Box?
body: <p>Here is my snippet:</p>
  
  <pre><code>private void btnBrowseCInv_Click(object sender, EventArgs e)
  {
    ofdBrowseVInv.Title = "Locate Customer Invoice File";
    ofdBrowseVInv.Filter = "Portable Document Format (*.pdf)|*.pdf|All Files (*.*)|*.*";
    ofdBrowseVInv.FileName = "";
    ofdBrowseVInv.FilterIndex = 0;
  
    ofdBrowseVInv.InitialDirectory = "";
  
    ofdBrowseVInv.CheckFileExists = true;
    ofdBrowseVInv.CheckPathExists = true;
  
    if (ofdBrowseVInv.ShowDialog() == System.Windows.Forms.DialogResult.OK)
    {
       //txtInvoicePathCInv.Text = ofdBrowseVInv... What property should i use?
    }
  }
  </code></pre>
  
  <p>As you see below, once a user pick a file and click open. I want the selected path to show on the pointed text box which is named "txtInvoicePathCInv". Any idea?</p>
  
  <p>I'm using Windows Application...</p>
  
  <p><a href="http://img708.imageshack.us/img708/54/99763211.jpg" rel="nofollow noreferrer">alt text http://img708.imageshack.us/img708/54/99763211.jpg</a></p>