How to Get the File name Using Open File Dialog

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OpenFileDialog dialog = new OpenFileDialog();
dialog.Filter = "jpg files |*.JPG";
dialog.InitialDirectory = "D:\\";
dialog.Title = "Select an Image file";
if (dialog.ShowDialog() == DialogResult.OK)
{
txtLocation.Text = dialog.FileName;
pbxImg.ImageLocation = txtLocation.Text;
txtPath.Text = Environment.CurrentDirectory;
}

pbxImg.Image.Save(Environment.CurrentDirectory+"----.Jpg", System.Drawing.Imaging.ImageFormat.Jpeg);

This is the Code I am Using..

I need to get just the File name from the Openfile dialog. and If I fillup that in the Blanks as a string. I will get the answer.

Posted 15-Feb-12 1:02am

adithya.hari

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**AmitGajjar** · Answer 1 · 2012-02-15T02:03:00

Hi,

Refer this link.

here is the sample code for getting selected file from OpenFileDialog,

C#

private void button1_Click(object sender, System.EventArgs e)
{
   if(openFileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK)
   {
      System.IO.StreamReader sr = new 
         System.IO.StreamReader(openFileDialog1.FileName);
      MessageBox.Show(sr.ReadToEnd());
      sr.Close();
   }
}

you can access selected file using openFileDialog.FileName.

hope this is what you want.

thanks
-amit.

santhosh-padamatinti · Answer 2 · 2012-02-15T02:11:00

C#

private void button1_Click(object sender, EventArgs e)
{

    if (openFileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK)
    {
        System.IO.StreamReader sr = new
           System.IO.StreamReader(openFileDialog1.FileName);
        string[] s=(openFileDialog1.FileName.ToString()).Split('\\');
        int count = s.Length;
        MessageBox.Show(s[count-1]);

        sr.Close();
    }

}

Raghubir_Sarkar · Answer 3 · 2014-05-22T22:19:00

Solution 3

System.IO.Path.GetFileName(Path as String)

Posted 22-May-14 22:19pm

Raghubir_Sarkar

Tanuj Vaja · Answer 4 · 2014-05-23T01:07:00

Solution 4

var FD = new System.Windows.Forms.OpenFileDialog();
string fileToOpen = "";
if (FD.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
fileToOpen = FD.FileName;
}

Here fileToOpen contains the filename.

Posted 23-May-14 1:07am

Tanuj Vaja

How to Get the File name Using Open File Dialog

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