How to make OpenFileLocation button

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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

C#

OpenFileDialog ofd = new OpenFileDialog();
            
            
            if (ofd.ShowDialog() == System.Windows.Forms.DialogResult.OK)
            {
                this.textBox1.Text = ofd.FileName;
                
            }

and I created this

C#

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

C#

if (ofd.ShowDialog() == DialogResult.OK)
            {
                textBox1.Text = OpenFileDialog;
            }

But , it didnt worked.

Posted 28-Nov-12 9:52am

Phoenix 1337

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Comments

Tamer Hatoum 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.

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Jason Gleim · Answer 1 · 2012-11-28T10:03:00

Solution 1

Use the FolderBrowserDialog class

C#

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

Posted 28-Nov-12 10:03am

Jason Gleim