我想要一个简单的正则表达式，可以允许太多，但不能拒绝有效的邮政编码。我这样做(输入是一个剥离/修剪的字符串):

/^([a-z0-9]\s*){5,8}$/i

这允许最短的邮政编码，如“L1 8JQ”和最长的邮政编码，如“OL14 5ET”。

因为它最多允许8个字符，如果没有空格，它也将允许不正确的8个字符邮政编码:“OL145ETX”。但是，这是一个简单的正则表达式，当它足够好的时候。

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

前半段邮政编码有效格式

[a - z] [a - z][0 - 9]的[a -ž] [a - z] [a - z] [0 - 9] [0 - 9] [a - z] [0 - 9] [0 - 9] [a - z] [a - z] [0 - 9] [a - z] [a - z]的[a -ž] [a - z][0 - 9]的[a -ž] [a - z] [0 - 9]

异常 位置1 - QVX未使用 位置2 -除GIR 0AA外，IJZ不使用 位置3 - AEHMNPRTVXY只使用 位置4 - ABEHMNPRVWXY

邮政编码的后半部分

[0 - 9] [a - z]的[a -ž]

异常 位置2+3 - CIKMOV未使用

记住，不是所有可能的代码都被使用了，所以这个列表是有效代码的必要条件，而不是充分条件。只是匹配所有有效代码的列表可能会更容易?

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。

这是谷歌在i18napis.appspot.com域名上的正则表达式:

GIR[ ]?0AA|((AB|AL|B|BA|BB|BD|BH|BL|BN|BR|BS|BT|BX|CA|CB|CF|CH|CM|CO|CR|CT|CV|CW|DA|DD|DE|DG|DH|DL|DN|DT|DY|E|EC|EH|EN|EX|FK|FY|G|GL|GY|GU|HA|HD|HG|HP|HR|HS|HU|HX|IG|IM|IP|IV|JE|KA|KT|KW|KY|L|LA|LD|LE|LL|LN|LS|LU|M|ME|MK|ML|N|NE|NG|NN|NP|NR|NW|OL|OX|PA|PE|PH|PL|PO|PR|RG|RH|RM|S|SA|SE|SG|SK|SL|SM|SN|SO|SP|SR|SS|ST|SW|SY|TA|TD|TF|TN|TQ|TR|TS|TW|UB|W|WA|WC|WD|WF|WN|WR|WS|WV|YO|ZE)(\d[\dA-Z]?[ ]?\d[ABD-HJLN-UW-Z]{2}))|BFPO[ ]?\d{1,4}