使用

[A-Z]?

使字母可选. {1}是多余的. (当然，您也可以写[A-Z]{0,1}，意思是一样的，但这就是?的用途.)

to make the letter optional. {1} is redundant. (Of course you could also write [A-Z]{0,1} which would mean the same, but that's what the ? is there for.)

您可以将正则表达式改进为

You could improve your regex to

^([0-9]{5})+\s+([A-Z]?)\s+([A-Z])([0-9]{3})([0-9]{3})([A-Z]{3})([A-Z]{3})\s+([A-Z])[0-9]{3}([0-9]{4})([0-9]{2})([0-9]{2})

而且，由于在大多数正则表达式中，\d与[0-9]相同:

And, since in most regex dialects, \d is the same as [0-9]:

^(\d{5})+\s+([A-Z]?)\s+([A-Z])(\d{3})(\d{3})([A-Z]{3})([A-Z]{3})\s+([A-Z])\d{3}(\d{4})(\d{2})(\d{2})

但是:您真的需要11个单独的捕获组吗?如果是这样，为什么不捕获倒数第四组数字呢?

But: do you really need 11 separate capturing groups? And if so, why don't you capture the fourth-to-last group of digits?