通过经验测试和观察，以及https://en.wikipedia.org/wiki/Postcodes_in_the_United_Kingdom#Validation的确认，以下是我的Python正则表达式版本，可以正确地解析和验证英国邮政编码:

UK_POSTCODE_REGEX = r ' (? P < postcode_area > [a - z] {1,2}) (? P <区> (?:[0 - 9]{1,2})| (?:[0 - 9][a - z])) (? P <部门> [0 - 9])(? P <邮编> [a - z]{2})”

这个正则表达式很简单，并且有捕获组。它不包括所有合法的英国邮政编码的验证，而只考虑字母与数字的位置。

下面是我在代码中如何使用它:

@dataclass
class UKPostcode:
    postcode_area: str
    district: str
    sector: int
    postcode: str

    # https://en.wikipedia.org/wiki/Postcodes_in_the_United_Kingdom#Validation
    # Original author of this regex: @jontsai
    # NOTE TO FUTURE DEVELOPER:
    # Verified through empirical testing and observation, as well as confirming with the Wiki article
    # If this regex fails to capture all valid UK postcodes, then I apologize, for I am only human.
    UK_POSTCODE_REGEX = r'(?P<postcode_area>[A-Z]{1,2})(?P<district>(?:[0-9]{1,2})|(?:[0-9][A-Z]))(?P<sector>[0-9])(?P<postcode>[A-Z]{2})'

    @classmethod
    def from_postcode(cls, postcode):
        """Parses a string into a UKPostcode

        Returns a UKPostcode or None
        """
        m = re.match(cls.UK_POSTCODE_REGEX, postcode.replace(' ', ''))

        if m:
            uk_postcode = UKPostcode(
                postcode_area=m.group('postcode_area'),
                district=m.group('district'),
                sector=m.group('sector'),
                postcode=m.group('postcode')
            )
        else:
            uk_postcode = None

        return uk_postcode


def parse_uk_postcode(postcode):
    """Wrapper for UKPostcode.from_postcode
    """
    uk_postcode = UKPostcode.from_postcode(postcode)
    return uk_postcode

下面是单元测试:

@pytest.mark.parametrize(
    'postcode, expected', [
        # https://en.wikipedia.org/wiki/Postcodes_in_the_United_Kingdom#Validation
        (
            'EC1A1BB',
            UKPostcode(
                postcode_area='EC',
                district='1A',
                sector='1',
                postcode='BB'
            ),
        ),
        (
            'W1A0AX',
            UKPostcode(
                postcode_area='W',
                district='1A',
                sector='0',
                postcode='AX'
            ),
        ),
        (
            'M11AE',
            UKPostcode(
                postcode_area='M',
                district='1',
                sector='1',
                postcode='AE'
            ),
        ),
        (
            'B338TH',
            UKPostcode(
                postcode_area='B',
                district='33',
                sector='8',
                postcode='TH'
            )
        ),
        (
            'CR26XH',
            UKPostcode(
                postcode_area='CR',
                district='2',
                sector='6',
                postcode='XH'
            )
        ),
        (
            'DN551PT',
            UKPostcode(
                postcode_area='DN',
                district='55',
                sector='1',
                postcode='PT'
            )
        )
    ]
)
def test_parse_uk_postcode(postcode, expected):
    uk_postcode = parse_uk_postcode(postcode)
    assert(uk_postcode == expected)

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。

我建议你看看英国政府的邮政编码数据标准[链接现在死了;XML的存档，参见维基百科的讨论]。这里有关于数据的简要描述，附带的xml模式提供了一个正则表达式。这可能不是你想要的，但会是一个很好的起点。RegEx与XML略有不同，因为给定的定义允许在格式A9A 9AA中第三个位置的P字符。

英国政府提供的正则表达式为:

([Gg][Ii][Rr] 0[Aa]{2})|((([A-Za-z][0-9]{1,2})|(([A-Za-z][A-Ha-hJ-Yj-y][0-9]{1,2})|(([A-Za-z][0-9][A-Za-z])|([A-Za-z][A-Ha-hJ-Yj-y][0-9][A-Za-z]?))))\s?[0-9][A-Za-z]{2})

正如维基百科讨论中指出的那样，这将允许一些非真实的邮政编码(例如以AA, ZY开头的邮政编码)，并且它们确实提供了一个更严格的测试，您可以尝试一下。

我有英国邮政编码验证的正则表达式。

这是适用于所有类型的邮政编码，无论是内部或外部

^((([A-PR-UWYZ][0-9])|([A-PR-UWYZ][0-9][0-9])|([A-PR-UWYZ][A-HK-Y][0-9])|([A-PR-UWYZ][A-HK-Y][0-9][0-9])|([A-PR-UWYZ][0-9][A-HJKSTUW])|([A-PR-UWYZ][A-HK-Y][0-9][ABEHMNPRVWXY]))) || ^((GIR)[ ]?(0AA))$|^(([A-PR-UWYZ][0-9])[ ]?([0-9][ABD-HJLNPQ-UW-Z]{0,2}))$|^(([A-PR-UWYZ][0-9][0-9])[ ]?([0-9][ABD-HJLNPQ-UW-Z]{0,2}))$|^(([A-PR-UWYZ][A-HK-Y0-9][0-9])[ ]?([0-9][ABD-HJLNPQ-UW-Z]{0,2}))$|^(([A-PR-UWYZ][A-HK-Y0-9][0-9][0-9])[ ]?([0-9][ABD-HJLNPQ-UW-Z]{0,2}))$|^(([A-PR-UWYZ][0-9][A-HJKS-UW0-9])[ ]?([0-9][ABD-HJLNPQ-UW-Z]{0,2}))$|^(([A-PR-UWYZ][A-HK-Y0-9][0-9][ABEHMNPRVWXY0-9])[ ]?([0-9][ABD-HJLNPQ-UW-Z]{0,2}))$

这适用于所有类型的格式。

例子:

Ab10 -------------------->仅为外部邮政编码 A1 1 aa ------------------> (内部和外部)邮政编码的组合 WC2A --------------------> 外

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。