How to open a file in VB6 using app.Path?

**Skate Bart** · December 24th, 2012, 07:59 AM

When the button is clicked, I want the file to be opened.
The file will be in the same directory as the visual basic project.
Therefore, I need the code to open the file using "app.Path".
Any solutions?

**dglienna** · December 24th, 2012, 11:42 AM

What kind of file are we talking about?

**Skate Bart** · December 24th, 2012, 11:47 AM

Originally Posted by dglienna

What kind of file are we talking about?

it's an XPS file (.xps).
When the button is clicked, I want that file to be opened in windows.

**Cereal Killer** · December 24th, 2012, 01:25 PM

You must get the full path of the file, like this

App.Path & "\" & <name.ext>

Once you have it, you can call the ShellExecute API function passing as param the string you got before.

**dglienna** · December 25th, 2012, 06:41 PM

And OPEN will open the text file, but not decode the XML inside of it.

http://www.fileinfo.com/extension/xps

**Cereal Killer** · December 26th, 2012, 04:37 AM

Originally Posted by dglienna

And OPEN will open the text file, but not decode the XML inside of it.

The OPEN argument is not needed and you can also leave a blank string, because the ShellExecute has the open command as default one. See a guide on this API, that provided by AllAPI.net is a very good one.

Enjoy!