CODEWITHSUNDEEP
c#.netopenfiledialog
AndrewD
AndrewD

Reputation: 75

C# OpenFileDialog for getting a user-selected output path for a Windows form?

Good morning,

I'm trying to figure out how to properly use the OpenFileDialog function in C# to allow a user to select a desired output folder. Right now I have a button and a textbox on my windows form. The user would hit the button, and that would open up the dialog GUI at run-time to allow the user to navigate to the the output location, then hit ok. Then they should have the confirmation of their selection by having the path displayed in the textbox.

The code I have right now is as follows: 

      private void button1_Click_3(object sender, EventArgs e)
    {
        OpenFileDialog OutputFilePath = new OpenFileDialog();

        string OutputString = OutputFilePath.FileName;
        FilePathBox.Text = OutputString;

    }

It compiles just fine, but when I hit the button, it doesn't bring up the file dialog box.

I'm sure it's something simple that I'm not seeing?

Thanks in advance!

~Andrew

Upvotes: 0

Views: 1416

Answers (2)

Chris L
Chris L

Reputation: 2292

You need to use ShowDialog()

private void button1_Click_3(object sender, EventArgs e)
{
    using(OpenFileDialog OutputFilePath = new OpenFileDialog())
    {
         if(OutputFilePath.ShowDialog() == DialogResult.OK)
         {
             string OutputString = OutputFilePath.FileName;
             FilePathBox.Text = OutputString;
         }
    }

}

Wrapping the dialog in a using clause will ensure the form is garbage collected.

Upvotes: 2

ASh
ASh

Reputation: 35720

you need to show Dialog and then check DialogResult because user can click Cancel

OpenFileDialog OutputFilePath = new OpenFileDialog();
var res = OutputFilePath.ShowDialog();
if (res == DialogResult.OK)
{
    string OutputString = OutputFilePath.FileName;
    FilePathBox.Text = OutputString;
}

Upvotes: 4

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