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

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?

这个允许两边有空格和制表符，以防你不想验证失败，然后在另一边修剪它。

^\s*(([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])))) {0,1}[0-9][A-Za-z]{2})\s*$)

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)

根据皇家邮政的程序员指南，检查邮政编码是否为有效格式:

          |----------------------------outward code------------------------------| |------inward code-----|
#special↓       α1        α2    AAN  AANA      AANN      AN    ANN    ANA (α3)        N         AA
^(GIR 0AA|[A-PR-UWYZ]([A-HK-Y]([0-9][A-Z]?|[1-9][0-9])|[1-9]([0-9]|[A-HJKPSTUW])?) [0-9][ABD-HJLNP-UW-Z]{2})$

uk上的所有邮编都匹配，除了那些不再使用的邮编。

增加一个?空格后，使用不区分大小写的匹配来回答这个问题:

'se50eg'.match(/^(GIR 0AA|[A-PR-UWYZ]([A-HK-Y]([0-9][A-Z]?|[1-9][0-9])|[1-9]([0-9]|[A-HJKPSTUW])?) ?[0-9][ABD-HJLNP-UW-Z]{2})$/ig);
Array [ "se50eg" ]