Get Relative Path to Linked Resource in C#

Question

I have a number of linked resources (text files in this case) in a C# project I am working on. I know the resources are linked via relative path in the Resources file. Is there any way to get that relative path from within the C# code?

If there isn't, what I am trying to do is to pass that resources file to an external application as a commandline argument. If I can't do it the first way (which would obviously be the easiest) any other suggestions would be much appreciated. Thanks!

Answer 1

I think you can get the resource, save it to a temporary folder and then supply the path to your command line external application, this is a quick and dirty way, but that's the main idea:

var assembly = System.Reflection.Assembly.GetExecutingAssembly();
var stream = assembly.GetManifestResourceStream("Resources.yourFile.txt");
var tempPath = Path.GetTempPath();
File.Save(stream, tempPath);

// use your tempPath here

Answer 2

Afte compilation the resource files are part of a dll. So they do not exist anymore in the file system you distribute. If you need to have the resource files as a ressource then Hueso's solution is the only way of passing the files on to an external application.

However, if you don't mind the files being accessible by the user you can set the BuildAction to Content and set the Copy to Output Directory to either Copy if newer or Copy always. The files are then copied into the output directory and can be accessed from there. If the files are in sub folders these subfolders are also copied into the output directory.

Edit:

For the various build actions and their effect see:

• http://stackoverflow.com/questions/145752/what-are-the-various-build-action-settings-in-vs-net-project-properties-and-wha

Answer 3

Theese 3 lines takes your resourcefile and output to byte array, then write to a file on disk.

byte[] byteArray = new byte[Project.Properties.Resources.File.Length]; Project.Properties.Resources.File.CopyTo(byteArray, 0);

File.WriteAllBytes("Filepath", byteArray);