根据维基百科的表格

这种模式适用于所有情况

(?:[A-Za-z]\d ?\d[A-Za-z]{2})|(?:[A-Za-z][A-Za-z\d]\d ?\d[A-Za-z]{2})|(?:[A-Za-z]{2}\d{2} ?\d[A-Za-z]{2})|(?:[A-Za-z]\d[A-Za-z] ?\d[A-Za-z]{2})|(?:[A-Za-z]{2}\d[A-Za-z] ?\d[A-Za-z]{2})

当在Android / Java上使用它时，使用\\d

接受的答案反映了皇家邮政给出的规则，尽管正则表达式中有一个拼写错误。这个错字似乎在gov.uk网站上也有(就像在XML存档页面中一样)。

在格式A9A 9AA中，规则允许在第三个位置出现P字符，而正则表达式不允许这样。正确的正则表达式应该是:

(GIR 0AA)|((([A-Z-[QVX]][0-9][0-9]?)|(([A-Z-[QVX]][A-Z-[IJZ]][0-9][0-9]?)|(([A-Z-[QVX]][0-9][A-HJKPSTUW])|([A-Z-[QVX]][A-Z-[IJZ]][0-9][ABEHMNPRVWXY])))) [0-9][A-Z-[CIKMOV]]{2}) 

将其缩短为以下正则表达式(使用Perl/Ruby语法):

(GIR 0AA)|([A-PR-UWYZ](([0-9]([0-9A-HJKPSTUW])?)|([A-HK-Y][0-9]([0-9ABEHMNPRVWXY])?))\s?[0-9][ABD-HJLNP-UW-Z]{2})

它还在第一个和第二个块之间包含一个可选的空格。

下面是一个基于链接到marcj答案的文档中指定格式的正则表达式:

/^[A-Z]{1,2}[0-9][0-9A-Z]? ?[0-9][A-Z]{2}$/

这和规格之间的唯一区别是，根据规格，最后两个字符不能在[CIKMOV]中。

编辑: 下面是另一个测试末尾字符限制的版本。

/^[A-Z]{1,2}[0-9][0-9A-Z]? ?[0-9][A-BD-HJLNP-UW-Z]{2}$/

看看本页的python代码:

http://www.brunningonline.net/simon/blog/archives/001292.html

I've got some postcode parsing to do. The requirement is pretty simple; I have to parse a postcode into an outcode and (optional) incode. The good new is that I don't have to perform any validation - I just have to chop up what I've been provided with in a vaguely intelligent manner. I can't assume much about my import in terms of formatting, i.e. case and embedded spaces. But this isn't the bad news; the bad news is that I have to do it all in RPG. :-( Nevertheless, I threw a little Python function together to clarify my thinking.

我用它来处理邮政编码。

我们得到了一个说明:

UK postcodes must be in one of the following forms (with one exception, see below): 
    § A9 9AA 
    § A99 9AA
    § AA9 9AA
    § AA99 9AA
    § A9A 9AA
    § AA9A 9AA
where A represents an alphabetic character and 9 represents a numeric character.
Additional rules apply to alphabetic characters, as follows:
    § The character in position 1 may not be Q, V or X
    § The character in position 2 may not be I, J or Z
    § The character in position 3 may not be I, L, M, N, O, P, Q, R, V, X, Y or Z
    § The character in position 4 may not be C, D, F, G, I, J, K, L, O, Q, S, T, U or Z
    § The characters in the rightmost two positions may not be C, I, K, M, O or V
The one exception that does not follow these general rules is the postcode "GIR 0AA", which is a special valid postcode.

我们想出了这个:

/^([A-PR-UWYZ][A-HK-Y0-9](?:[A-HJKS-UW0-9][ABEHMNPRV-Y0-9]?)?\s*[0-9][ABD-HJLNP-UW-Z]{2}|GIR\s*0AA)$/i

但是注意-这允许组之间有任意数量的空格。

我今天做了英国邮政编码验证的正则表达式，据我所知，它适用于所有的英国邮政编码，如果你放一个空格或如果你不放。

^((([a-zA-Z][0-9])|([a-zA-Z][0-9]{2})|([a-zA-Z]{2}[0-9])|([a-zA-Z]{2}[0-9]{2})|([A-Za-z][0-9][a-zA-Z])|([a-zA-Z]{2}[0-9][a-zA-Z]))(\s*[0-9][a-zA-Z]{2})$)

如果有什么格式没有涵盖，请告诉我