SQLDiagRunner.Runner.GetOutputFilepath C# (CSharp) Метод

GetOutputFilepath() приватный статический Метод

private static GetOutputFilepath ( string outputFolder, string servername, string dateString ) : string
outputFolder string
servername string
dateString string
Результат string
        private static string GetOutputFilepath(string outputFolder, string servername, string dateString)
        {
            string ret = Directory.Exists(outputFolder)
                             ? Path.Combine(outputFolder, dateString + servername.ReplaceInvalidFilenameChars("_") + ".xlsx")
                             : outputFolder;

            return ret;
        }