ICSharpCode.ILSpy.CSharpLanguage.GetFileNameForResource C# (CSharp) Метод

		string GetFileNameForResource(string fullName, HashSet<string> directories)
		{
			string[] splitName = fullName.Split('.');
			string fileName = cleanupName(fullName);
			for (int i = splitName.Length - 1; i > 0; i--) {
				string ns = string.Join(".", splitName, 0, i);
				if (directories.Contains(ns)) {
					string name = string.Join(".", splitName, i, splitName.Length - i);
					fileName = Path.Combine(ns, cleanupName(name));
					break;
				}
			}
			return fileName;
		}
		#endregion