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Hey
 
I want to know how to build OpenFileLocation button.
I mean , after I select browse , ofd shows up. I select the .exe or whatever , then it prints to textbox1.text
OpenFileDialog ofd = new OpenFileDialog();
            
            
            if (ofd.ShowDialog() == System.Windows.Forms.DialogResult.OK)
            {
                this.textBox1.Text = ofd.FileName;
                
            }
and I created this
Process.Start(textbox1.Text)
but , I want to create a button to open the folder which locates the exe i selected from the browse button.
Like , Open file location button
I tried this
if (ofd.ShowDialog() == DialogResult.OK)
            {
                textBox1.Text = OpenFileDialog;
            }
But , it didnt worked.
Posted 28-Nov-12 9:52am
Comments
Tamer Hatoum at 28-Nov-12 16:00pm
   
if you mean that you want to locate an .exe file then you want to start the chosen app. then you have to start the process by the file path not only the file name.

1 solution

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Solution 1

Use the FolderBrowserDialog class
 

if (ofd.ShowDialog() == DialogResult.OK)
{
   FolderBrowserDialog dialog = new FolderBrowserDialog();
   dialog.SelectedPath = Path.GetDirectoryName(ofd.FileName);
   dialog.ShowDialog();
   textBox1.Text = dialog.SelectedPath;
}
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