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?
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
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)
其他回答
看起来我们将使用^(GIR ?0AA|[a - pr - uwyz]([0-9]{1,2}|([a - hk - y][0-9]([0-9ABEHMNPRV-Y])?)|[0-9][a - hjkps - uw])? [0-9][ABD-HJLNP-UW-Z]{2})$,这是上面Minglis建议的略有修改的版本。
然而,我们将不得不调查到底是什么规则,因为上面列出的各种解决方案似乎适用于不同的规则,哪些字母是允许的。
经过一番研究,我们找到了更多的信息。显然,“govtalk.gov.uk”上的一个页面会指向邮政编码规范govtalk-postcodes。它指向XML schema中的一个XML模式,该模式提供了邮政编码规则的“伪正则表达式”语句。
我们用它做了一些修改,得到了下面的表达式:
^((GIR &0AA)|((([A-PR-UWYZ][A-HK-Y]?[0-9][0-9]?)|(([A-PR-UWYZ][0-9][A-HJKSTUW])|([A-PR-UWYZ][A-HK-Y][0-9][ABEHMNPRV-Y]))) &[0-9][ABD-HJLNP-UW-Z]{2}))$
这使得空格是可选的,但限制您只能使用一个空格(将'&'替换为'{0,}表示无限空格)。它假定所有文本都必须是大写的。
如果你想要允许小写,任意数量的空格,使用:
^(([gG][iI][rR] {0,}0[aA]{2})|((([a-pr-uwyzA-PR-UWYZ][a-hk-yA-HK-Y]?[0-9][0-9]?)|(([a-pr-uwyzA-PR-UWYZ][0-9][a-hjkstuwA-HJKSTUW])|([a-pr-uwyzA-PR-UWYZ][a-hk-yA-HK-Y][0-9][abehmnprv-yABEHMNPRV-Y]))) {0,}[0-9][abd-hjlnp-uw-zABD-HJLNP-UW-Z]{2}))$
这并不包括海外领土,只是强制执行格式,而不是不同地区的存在。它基于以下规则:
可接受以下格式:
“秋天” A9 9 zz A99 9 zz AB9 9 zz AB99 9 zz A9C 9 zz AD9E 9 zz
地点:
9可以是任何一位数。 A可以是除Q、V或X之外的任何字母。 B可以是除I、J或Z之外的任何字母。 C可以是除I、L、M、N、O、P、Q、R、V、X、Y或Z之外的任何字母。 D可以是除I、J或Z之外的任何字母。 E可以是A, B, E, H, M, N, P, R, V, W, X或Y中的任意一个。 Z可以是C、I、K、M、O或V之外的任何字母。
最好的祝愿
科林
在这个列表中添加一个更实用的正则表达式,允许用户输入一个空字符串:
^$|^(([gG][iI][rR] {0,}0[aA]{2})|((([a-pr-uwyzA-PR-UWYZ][a-hk-yA-HK-Y]?[0-9][0-9]?)|(([a-pr-uwyzA-PR-UWYZ][0-9][a-hjkstuwA-HJKSTUW])|([a-pr-uwyzA-PR-UWYZ][a-hk-yA-HK-Y][0-9][abehmnprv-yABEHMNPRV-Y]))) {0,1}[0-9][abd-hjlnp-uw-zABD-HJLNP-UW-Z]{2}))$
这个正则表达式允许大写字母和小写字母,中间有可选的空格
从软件开发人员的角度来看,这个正则表达式对于地址可能是可选的软件很有用。例如,如果用户不想提供他们的地址详细信息
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
^([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
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。