在Visual Studio中使用嵌入式资源的代码或命令

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();
        }
    }
}