Can I open a resourcepath with File.Open in C#

Question

I try to open a resourcefile in C# using this line, I can't see why this not working. I have tried to use build action = "embedded resource" and "none". I have tried all ways of copy to output directory. Do I need some special way of opening up resourcefiles to fileStream with c#?

    public bool openFileStream()
    {
        FileStream mFileStream;
        String[] s=System.Reflection.Assembly.GetExecutingAssembly().GetManifestResourceNames();
        String mFilename =null;
        for (int i = 0; i < s.Length;i++ )
        {
            s[i] = s[i].Replace(".", "\\");

            if (s[i].Contains("sju_tre_g_i"))
            {
                s[i] = s[i].Substring(0, s[i].Length - 4);
                s[i] = s[i] + ".wav";
                mFilename = s[i];

            }
            System.Diagnostics.Debug.WriteLine(i + "___Outputname:" + mFilename);
        }
        System.Diagnostics.Debug.WriteLine(mFilename);
        try{
            mFileStream = File.Open(mFilename,FileMode.Open );
            return true;
        }
        catch(Exception)
        {
            return false;
        }
    }

EDIT not duplicate In the duplicate there is just to open a text-file. It is the adress to use to FileStream mFileStream; mFileStream=File.Open("WindowsFormsApplication2\\Resources\\‌​sju_tre_g_i.wav",Fil‌​eMode.Open );

Answer 1

If you want to open an wav file from the resouces you can do this.

using (var steam = Properties.Resources.sju_tre_g_i)
{
     // do something       
}

Just be aware that this isn't a FileStream but an UnmanagedMemoryStream . (But as the ressource library is loaded into memory anyway it shouldn't make a difference?)

---

If you want to ensure a directory exists and you want to open a file when it exists and create it when it doesn't you can do the following.

var path = @"C:\mypath\myfile.txt";
Directory.CreateDirectory(path);
using (var stream = File.Open(path, FileMode.OpenOrCreate))
{
     // do something            
}

You can also use an relative path if you like.