Member Avatar for WJSwan

Hi,
I have an application where I would like to add the following:

When the user clicks on a button the FileOpenDialog must activate and allow the user to select an EXE file. When the user clicks OK, the selected EXE should execute.

I am new to C# so I need some help here, PLEASE!

Regards,
Wimpie

  • 3 Contributors
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  • 7 Hours Discussion Span
  • Latest Post Latest Post by Ramy Mahrous

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Member Avatar for Ramy Mahrous 
if (openFileDialog1.ShowDialog() == DialogResult.OK)
            {
                System.Diagnostics.Process.Start(openFileDialog1.FileName);
            }
Member Avatar for LizR

WJSwan, if you look in your helpfile you will find examples too, it shows how to use other properties of the openfile dialog too.

Member Avatar for WJSwan

Hi RamyMahrous and LizR,
Thanks a stack for your prompt responses. My application is now working like a charm!!

Regards,
Wimpie

Member Avatar for Ramy Mahrous

Marke this thread as solved! please.

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