我有一个COM pressed字符串值，我是从导入文件中提取。我需要格式化这个成一个包号，格式如下： ## - ## - ## - ### - ### 。因此，因此，字符串410151000640应该成为41-01-51-000-640。我可以用下面的code做到这一点：

 的String.Format（{0：##  -  ##  -  ##  -  ###  -  ###}，Convert.ToInt64（410151000640））;
 

不过，该字符串可能不是所有的数字;它可以有一个或两个字母在那里，因此，换算为int将失败。有没有办法做到这一点的一个字符串，所以每个字符，不管它是一个数字或字母，将适合的格式是否正确？

  Regex.Replace（410151000640，@^（。{2}）（{2}）（ 。{2}）（{3}）（{3}）$，$ 1- $ $ 2- 3- 4- $ $ 5）;
 

还是略短版

  Regex.Replace（410151000640，@^（..）（..）（..）（...）（...）$，$ 1 -  $ 2- $ 3- $ 4- $ 5）;
 

I have a compressed string value I'm extracting from an import file. I need to format this into a parcel number, which is formatted as follows: ##-##-##-###-###. So therefore, the string "410151000640" should become "41-01-51-000-640". I can do this with the following code:

String.Format("{0:##-##-##-###-###}", Convert.ToInt64("410151000640"));

However, The string may not be all numbers; it could have a letter or two in there, and thus the conversion to the int will fail. Is there a way to do this on a string so every character, regardless of if it is a number or letter, will fit into the format correctly?

Regex.Replace("410151000640", @"^(.{2})(.{2})(.{2})(.{3})(.{3})$", "$1-$2-$3-$4-$5");

Or the slightly shorter version

Regex.Replace("410151000640", @"^(..)(..)(..)(...)(...)$", "$1-$2-$3-$4-$5");