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)

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。

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?

虽然这里有很多答案，但我对其中任何一个都不满意。他们中的大多数只是简单地坏了，太复杂或只是坏了。

我看了@ctwheels的答案，我发现它非常具有解释性和正确性;我们必须为此感谢他。然而，对我来说，如此简单的事情又有太多的“数据”了。

幸运的是，我设法获得了一个数据库，其中仅包含英国的100多万个活动邮政编码，并编写了一个小型PowerShell脚本来测试和基准测试结果。

英国邮政编码规格:有效的邮政编码格式。

这是“我的”正则表达式:

^([a-zA-Z]{1,2}[a-zA-Z\d]{1,2})\s(\d[a-zA-Z]{2})$

简短，简单，甜蜜。即使是最没有经验的人也能明白发生了什么。

解释:

^ asserts position at start of a line
    1st Capturing Group ([a-zA-Z]{1,2}[a-zA-Z\d]{1,2})
        Match a single character present in the list below [a-zA-Z]
        {1,2} matches the previous token between 1 and 2 times, as many times as possible, giving back as needed (greedy)
        a-z matches a single character in the range between a (index 97) and z (index 122) (case sensitive)
        A-Z matches a single character in the range between A (index 65) and Z (index 90) (case sensitive)
        Match a single character present in the list below [a-zA-Z\d]
        {1,2} matches the previous token between 1 and 2 times, as many times as possible, giving back as needed (greedy)
        a-z matches a single character in the range between a (index 97) and z (index 122) (case sensitive)
        A-Z matches a single character in the range between A (index 65) and Z (index 90) (case sensitive)
        \d matches a digit (equivalent to [0-9])
        \s matches any whitespace character (equivalent to [\r\n\t\f\v ])
    2nd Capturing Group (\d[a-zA-Z]{2})
        \d matches a digit (equivalent to [0-9])
        Match a single character present in the list below [a-zA-Z]
        {2} matches the previous token exactly 2 times
        a-z matches a single character in the range between a (index 97) and z (index 122) (case sensitive)
        A-Z matches a single character in the range between A (index 65) and Z (index 90) (case sensitive)
$ asserts position at the end of a line

结果(已核对邮编):

TOTAL OK: 1469193
TOTAL FAILED: 0
-------------------------------------------------------------------------
Days              : 0
Hours             : 0
Minutes           : 5
Seconds           : 22
Milliseconds      : 718
Ticks             : 3227185939
TotalDays         : 0.00373516891087963
TotalHours        : 0.0896440538611111
TotalMinutes      : 5.37864323166667
TotalSeconds      : 322.7185939
TotalMilliseconds : 322718.5939

基本规则:

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

英国的邮政编码(或称为邮政编码)由5到7个字母数字字符组成，中间用空格隔开。规定哪些角色可以出现在特定位置的规则相当复杂，而且充满了例外。因此，刚才显示的正则表达式遵循基本规则。

完整的规则:

如果你需要一个以牺牲可读性为代价的正则表达式来满足所有的邮政编码规则，这里你可以:

^(?:(?:[A-PR-UWYZ][0-9]{1,2}|[A-PR-UWYZ][A-HK-Y][0-9]{1,2}|[A-PR-UWYZ][0-9][A-HJKSTUW]|[A-PR-UWYZ][A-HK-Y][0-9][ABEHMNPRV-Y]) [0-9][ABD-HJLNP-UW-Z]{2}|GIR 0AA)$

来源:https://www.safaribooksonline.com/library/view/regular-expressions-cookbook/9781449327453/ch04s16.html

在我们的客户数据库中进行了测试，似乎非常准确。

^([A-PR-UWYZ0-9][A-HK-Y0-9][AEHMNPRTVXY0-9]?[ABEHMNPRVWXY0-9]? {1,2}[0-9][ABD-HJLN-UW-Z]{2}|GIR 0AA)$

Regular expression to match valid UK postcodes. In the UK postal system not all letters are used in all positions (the same with vehicle registration plates) and there are various rules to govern this. This regex takes into account those rules. Details of the rules: First half of postcode Valid formats [A-Z][A-Z][0-9][A-Z] [A-Z][A-Z][0-9][0-9] [A-Z][0-9][0-9] [A-Z][A-Z][0-9] [A-Z][A-Z][A-Z] [A-Z][0-9][A-Z] [A-Z][0-9] Exceptions Position - First. Contraint - QVX not used Position - Second. Contraint - IJZ not used except in GIR 0AA Position - Third. Constraint - AEHMNPRTVXY only used Position - Forth. Contraint - ABEHMNPRVWXY Second half of postcode Valid formats [0-9][A-Z][A-Z] Exceptions Position - Second and Third. Contraint - CIKMOV not used

http://regexlib.com/REDetails.aspx?regexp_id=260