GUID获取16位19位22位的唯一字符串

根据GUID获取16位的唯一字符串	public static string GuidTo16String()     {         long i = 1;         foreach (byte b in Guid.NewGuid().ToByteArray())             i *= ((int)b + 1);         return string.Format("{0:x}", i - DateTime.Now.Ticks);     }
根据GUID获取19位的唯一数字序列	public static long GuidToLongID()     {         byte[] buffer = Guid.NewGuid().ToByteArray();         return BitConverter.ToInt64(buffer, 0);     }
生成22位唯一的数字 并发可用	public static string GenerateUniqueID()     {         System.Threading.Thread.Sleep(1); //保证yyyyMMddHHmmssffff唯一         Random d = new Random(BitConverter.ToInt32(Guid.NewGuid().ToByteArray(), 0));         string strUnique = DateTime.Now.ToString("yyyyMMddHHmmssffff") + d.Next(1000, 9999);         return strUnique;     }