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
下面是一个基于链接到marcj答案的文档中指定格式的正则表达式:
/^[A-Z]{1,2}[0-9][0-9A-Z]? ?[0-9][A-Z]{2}$/
这和规格之间的唯一区别是，根据规格，最后两个字符不能在[CIKMOV]中。
编辑: 下面是另一个测试末尾字符限制的版本。
/^[A-Z]{1,2}[0-9][0-9A-Z]? ?[0-9][A-BD-HJLNP-UW-Z]{2}$/
其他回答
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)
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.
邮政编码可能会发生变化，验证邮政编码的唯一真正方法是拥有完整的邮政编码列表，并查看它是否存在。
但是正则表达式很有用，因为它们:
是否易于使用和实现 是短暂的 都跑得很快 相当容易维护(与完整的邮政编码列表相比) 仍然捕获大多数输入错误
但是正则表达式往往很难维护，特别是对于那些一开始就没有想到它的人来说。所以它一定是:
尽量简单易懂 相对未来的证明
这意味着这个答案中的大多数正则表达式都不够好。例如，我可以看到[a - pr - uwyz][a - hk - y][0-9][ABEHMNPRV-Y]将匹配形式为AA1A的邮政编码区域-但如果添加了新的邮政编码区域，这将是一个令人头疼的问题，因为很难理解它匹配哪些邮政编码区域。
我还想让我的正则表达式匹配邮政编码的前半部分和后半部分。
所以我想到了这个:
(GIR(?=\s*0AA)|(?:[BEGLMNSW]|[A-Z]{2})[0-9](?:[0-9]|(?<=N1|E1|SE1|SW1|W1|NW1|EC[0-9]|WC[0-9])[A-HJ-NP-Z])?)\s*([0-9][ABD-HJLNP-UW-Z]{2})
在PCRE格式中，可以这样写:
/^
( GIR(?=\s*0AA) # Match the special postcode "GIR 0AA"
|
(?:
[BEGLMNSW] | # There are 8 single-letter postcode areas
[A-Z]{2} # All other postcode areas have two letters
)
[0-9] # There is always at least one number after the postcode area
(?:
[0-9] # And an optional extra number
|
# Only certain postcode areas can have an extra letter after the number
(?<=N1|E1|SE1|SW1|W1|NW1|EC[0-9]|WC[0-9])
[A-HJ-NP-Z] # Possible letters here may change, but [IO] will never be used
)?
)
\s*
([0-9][ABD-HJLNP-UW-Z]{2}) # The last two letters cannot be [CIKMOV]
$/x
对我来说，这是尽可能多地验证之间的正确平衡，与此同时，未来的验证和易于维护。
我有英国邮政编码验证的正则表达式。
这是适用于所有类型的邮政编码，无论是内部或外部
^((([A-PR-UWYZ][0-9])|([A-PR-UWYZ][0-9][0-9])|([A-PR-UWYZ][A-HK-Y][0-9])|([A-PR-UWYZ][A-HK-Y][0-9][0-9])|([A-PR-UWYZ][0-9][A-HJKSTUW])|([A-PR-UWYZ][A-HK-Y][0-9][ABEHMNPRVWXY]))) || ^((GIR)[ ]?(0AA))$|^(([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
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
