c# - Code or command to use embedded resource in Visual Studio

Question

Can somebody provide me a starting point or code to access an embedded resource using C#?

I have successfully embedded a couple of batch files, scripts and CAD drawings which I would like to run the batch and copy the scripts and CAD files to a location specified in the batch file.

I'm struggling to find how to specify what the item is and set the path within the EXE. The below code is what I thought would work, but it failed and the others I found online all related to XML files.

System.Diagnostics.Process p = new System.Diagnostics.Process();
p.StartInfo.FileName = AppDomain.CurrentDomain.BaseDirectory + "\Batchfile.bat";
p.Start();

I honestly don't even know if I'm looking at the correct way to do this as this is my first time using either C# or Visual Studio.

Answer 1

Open Solution Explorer add files you want to embed. Right click on the files then click on Properties. In Properties window and change Build Action to Embedded Resource.

After that you should write the embedded resources to file in order to be able to run it.

using System;
using System.Reflection;
using System.IO;
using System.Diagnostics;

namespace YourProject
{
    public class MyClass
    {
        // Other Code...

        private void StartProcessWithFile()
        {
            var assembly = Assembly.GetExecutingAssembly();
            //Getting names of all embedded resources
            var allResourceNames = assembly.GetManifestResourceNames();
            //Selecting first one. 
            var resourceName = allResourceNames[0];
            var pathToFile = Path.GetDirectoryName(AppDomain.CurrentDomain.BaseDirectory) +
                              resourceName;

            using (var stream = assembly.GetManifestResourceStream(resourceName))
            using (var fileStream = File.Create(pathToFile))
            {
                stream.Seek(0, SeekOrigin.Begin);
                stream.CopyTo(fileStream);
            }

            var process = new Process();
            process.StartInfo.FileName = pathToFile;
            process.Start();
        }
    }
}