我在一个正则表达式后,将验证一个完整的复杂的英国邮政编码只在输入字符串。所有不常见的邮政编码形式必须包括以及通常。例如:

匹配

CW3 9不锈钢 SE5 0EG SE50EG Se5 0eg WC2H 7LT

不匹配

aWC2H 7LT WC2H 7LTa WC2H

我怎么解决这个问题?


当前回答

我一直在寻找一个英国邮政编码正则表达式的最后一天左右,无意中发现了这个线程。我尝试了上面的大部分建议,但没有一个对我有用,所以我想出了自己的正则表达式,据我所知,它捕获了截至1月13日的所有有效的英国邮政编码(根据皇家邮政的最新文献)。

The regex and some simple postcode checking PHP code is posted below. NOTE:- It allows for lower or uppercase postcodes and the GIR 0AA anomaly but to deal with the, more than likely, presence of a space in the middle of an entered postcode it also makes use of a simple str_replace to remove the space before testing against the regex. Any discrepancies beyond that and the Royal Mail themselves don't even mention them in their literature (see http://www.royalmail.com/sites/default/files/docs/pdf/programmers_guide_edition_7_v5.pdf and start reading from page 17)!

注意:在皇家邮政自己的文献中(链接以上),第3和第4位的位置略有模糊,如果这些字符是字母,则例外。我直接联系了皇家邮政,用他们自己的话说,“AANA NAA格式的出境代码的第4个位置的信件没有例外,而第3个位置的例外只适用于ANA NAA格式的出境代码的最后一个字母。”直接从马嘴里说出来的!

<?php

    $postcoderegex = '/^([g][i][r][0][a][a])$|^((([a-pr-uwyz]{1}([0]|[1-9]\d?))|([a-pr-uwyz]{1}[a-hk-y]{1}([0]|[1-9]\d?))|([a-pr-uwyz]{1}[1-9][a-hjkps-uw]{1})|([a-pr-uwyz]{1}[a-hk-y]{1}[1-9][a-z]{1}))(\d[abd-hjlnp-uw-z]{2})?)$/i';

    $postcode2check = str_replace(' ','',$postcode2check);

    if (preg_match($postcoderegex, $postcode2check)) {

        echo "$postcode2check is a valid postcode<br>";

    } else {

        echo "$postcode2check is not a valid postcode<br>";

    }

?>

我希望它能帮助其他遇到这条线索寻找解决方案的人。

其他回答

不存在能够验证邮政编码的综合英国邮政编码正则表达式。您可以使用正则表达式检查邮政编码的格式是否正确;并不是真的存在。

邮政编码非常复杂,而且不断变化。例如,对于每个邮政编码区域,出码W1没有,也可能永远没有1到99之间的每个数字。

你不能指望当前的东西永远都是真的。举个例子,1990年,邮局认为阿伯丁有点拥挤了。他们在AB1-5的末尾加了一个0,使它成为AB10-50,然后在这些之间创建了一些邮政编码。

每当建立一条新街道时,就会创建一个新的邮政编码。这是获得建筑许可的过程的一部分;地方当局有义务与邮局保持更新(并不是说他们都这样做)。

此外,正如许多其他用户指出的那样,还有一些特殊的邮政编码,如Girobank, GIR 0AA,以及给圣诞老人的信件,SAN TA1 -你可能不想在那里张贴任何东西,但似乎没有任何其他答案。

然后,还有BFPO的邮政编码,现在正在改为更标准的格式。两种格式都是有效的。最后,还有海外领土来源维基百科。

+----------+----------------------------------------------+
| Postcode |                   Location                   |
+----------+----------------------------------------------+
| AI-2640  | Anguilla                                     |
| ASCN 1ZZ | Ascension Island                             |
| STHL 1ZZ | Saint Helena                                 |
| TDCU 1ZZ | Tristan da Cunha                             |
| BBND 1ZZ | British Indian Ocean Territory               |
| BIQQ 1ZZ | British Antarctic Territory                  |
| FIQQ 1ZZ | Falkland Islands                             |
| GX11 1AA | Gibraltar                                    |
| PCRN 1ZZ | Pitcairn Islands                             |
| SIQQ 1ZZ | South Georgia and the South Sandwich Islands |
| TKCA 1ZZ | Turks and Caicos Islands                     |
+----------+----------------------------------------------+

接下来,你必须考虑到英国将其邮政编码系统“输出”到世界上许多地方。任何验证“英国”邮政编码的程序也将验证许多其他国家的邮政编码。

如果您想验证英国邮政编码,最安全的方法是使用当前邮政编码的查找。有很多选择:

Ordnance Survey releases Code-Point Open under an open data licence. It'll be very slightly behind the times but it's free. This will (probably - I can't remember) not include Northern Irish data as the Ordnance Survey has no remit there. Mapping in Northern Ireland is conducted by the Ordnance Survey of Northern Ireland and they have their, separate, paid-for, Pointer product. You could use this and append the few that aren't covered fairly easily. Royal Mail releases the Postcode Address File (PAF), this includes BFPO which I'm not sure Code-Point Open does. It's updated regularly but costs money (and they can be downright mean about it sometimes). PAF includes the full address rather than just postcodes and comes with its own Programmers Guide. The Open Data User Group (ODUG) is currently lobbying to have PAF released for free, here's a description of their position. Lastly, there's AddressBase. This is a collaboration between Ordnance Survey, Local Authorities, Royal Mail and a matching company to create a definitive directory of all information about all UK addresses (they've been fairly successful as well). It's paid-for but if you're working with a Local Authority, government department, or government service it's free for them to use. There's a lot more information than just postcodes included.

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。

上面的一些正则表达式有点限制性。请注意真正的邮政编码:“W1K 7AA”将失败,因为上面的“位置3 - AEHMNPRTVXY仅使用”规则将不允许“K”。

正则表达式:

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

似乎更准确一点,请参阅维基百科上题为“英国的邮政编码”的文章。

注意,这个正则表达式只要求大写字符。

更大的问题是,您是限制用户输入,只允许实际存在的邮政编码,还是只是试图阻止用户在表单字段中输入完全的垃圾。正确匹配每一个可能的邮政编码,并在未来校对,是一个更难的难题,除非你是HMRC,否则可能不值得这么做。

^([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

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)