Odtnhj

I want to extract the resources "cygz.dll" to the current directory where the program started right after the form is loaded.

Here is my code.

private static void extract(string nameSpace, string outDirectory, string internalFilePath, string resourceName)
 
 Assembly assembly = Assembly.GetCallingAssembly();

 using (Stream s = assembly.GetManifestResourceStream(nameSpace + "." + (internalFilePath == "" ? "" : internalFilePath + ".") + resourceName))

 using (BinaryReader r = new BinaryReader(s))

 using (FileStream fs = new FileStream(outDirectory + "\" + resourceName, FileMode.OpenOrCreate))

 using (BinaryWriter w = new BinaryWriter(fs))

 w.Write(r.ReadBytes((int)s.Length));

 

private void Form1_Load(object sender, EventArgs e)
 

 extract("myNamespace", "\TEMP", "resources", "cygz.dll");
 

Instead of "\TEMP", I want to extract "cygz.dll" to the current directory

If by "current directory" you mean the path for the executable file that started the application, then you can use Application.ExecutablePath like this:

System.IO.Path.GetDirectoryName(Application.ExecutablePath);

You can find more details here: https://docs.microsoft.com/en-us/dotnet/api/system.windows.forms.application.executablepath?view=netframework-4.7.2

This solved my problem

 private void Form1_Load(object sender, EventArgs e)


 string path = Environment.CurrentDirectory;

 extract("myNamespace", path, "resources", "cygz.dll");