OpenXML documents, how do you know which is which, when there is no extention

Question

What i have done for now, and which works is this:

private string DetermineOpenXML(string file)
    {

        try
        {
            SpreadsheetDocument doc = SpreadsheetDocument.Open(file, false);
            doc.Close();
            return ".xslx";
        }
        catch
        {
            try
            {
                WordprocessingDocument doc = WordprocessingDocument.Open(file, false);
                doc.Close();
                return ".docx";
            }
            catch
            {
                try
                {
                    PresentationDocument doc = PresentationDocument.Open(file, false);
                    doc.Close();
                    return ".pptx";
                }
                catch
                {
                    return string.Empty;
                }
            }
        }
    }

I think there must be a better way to see what kind of file it is other than just trial and error. The thing is, i am working on a small program, that finds out what file extension files should have. The reason i do this is because i have the files from a database, where they are saved sometimes without extension and other times with a wrong extension.

What i have done with these files is that i found out that all OpenXML documents share the same File signature: "50 4B 03 04 14 00 06 00", which is close to a signature of a zip file, and i can also open OpenXML files with a zip program and see its content. And maybe this is the solution i should go for, i was just hoping that it would be faster / easier to use OpenXML SDK and that it had a property or something that could check it for me.

Edit: I have added a answer, i would still like to see if there was a better solution, even though there it works for my current purpose. It does not take in for account that the extensions should have been template files.

Answer 1

I ended up using System.IO.Packaging instead.

private string anotherOpenXmlAttempt(string file)
    {
        string ext = string.Empty;
        Package package = Package.Open(file);
        if (package.PartExists(new Uri("/word/document.xml", UriKind.Relative)))
        {
            ext = ".docx";
        }
        else if (package.PartExists(new Uri("/xl/workbook.xml", UriKind.Relative)))
        {
            ext = ".xslx";
        }else if (package.PartExists(new Uri("/ppt/presentation.xml", UriKind.Relative)))
        {
            ext = ".pptx";
        }

        package.Close();
        return ext;
    }

havn't done any extensive testing, but have worked for my current files.

I will leave the question open in case someone has a nice solution.