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
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
其他回答
这是谷歌在i18napis.appspot.com域名上的正则表达式:
GIR[ ]?0AA|((AB|AL|B|BA|BB|BD|BH|BL|BN|BR|BS|BT|BX|CA|CB|CF|CH|CM|CO|CR|CT|CV|CW|DA|DD|DE|DG|DH|DL|DN|DT|DY|E|EC|EH|EN|EX|FK|FY|G|GL|GY|GU|HA|HD|HG|HP|HR|HS|HU|HX|IG|IM|IP|IV|JE|KA|KT|KW|KY|L|LA|LD|LE|LL|LN|LS|LU|M|ME|MK|ML|N|NE|NG|NN|NP|NR|NW|OL|OX|PA|PE|PH|PL|PO|PR|RG|RH|RM|S|SA|SE|SG|SK|SL|SM|SN|SO|SP|SR|SS|ST|SW|SY|TA|TD|TF|TN|TQ|TR|TS|TW|UB|W|WA|WC|WD|WF|WN|WR|WS|WV|YO|ZE)(\d[\dA-Z]?[ ]?\d[ABD-HJLN-UW-Z]{2}))|BFPO[ ]?\d{1,4}
我发现在几乎所有的变化和regex从批量转移pdf和什么是在维基百科网站上是这样的,特别是维基百科的regex是,需要有一个^后的第一个|(竖条)。我通过测试AA9A 9AA发现了这一点,因为否则A9A 9AA的格式检查将验证它。例如,检查应该无效的EC1D 1BB返回有效,因为C1D 1BB是有效的格式。
以下是我想出的一个好的正则表达式:
^([G][I][R] 0[A]{2})|^((([A-Z-[QVX]][0-9]{1,2})|([A-Z-[QVX]][A-HK-Y][0-9]{1,2})|([A-Z-[QVX]][0-9][ABCDEFGHJKPSTUW])|([A-Z-[QVX]][A-HK-Y][0-9][ABEHMNPRVWXY])) [0-9][A-Z-[CIKMOV]]{2})$
我们得到了一个说明:
UK postcodes must be in one of the following forms (with one exception, see below): § A9 9AA § A99 9AA § AA9 9AA § AA99 9AA § A9A 9AA § AA9A 9AA where A represents an alphabetic character and 9 represents a numeric character. Additional rules apply to alphabetic characters, as follows: § The character in position 1 may not be Q, V or X § The character in position 2 may not be I, J or Z § The character in position 3 may not be I, L, M, N, O, P, Q, R, V, X, Y or Z § The character in position 4 may not be C, D, F, G, I, J, K, L, O, Q, S, T, U or Z § The characters in the rightmost two positions may not be C, I, K, M, O or V The one exception that does not follow these general rules is the postcode "GIR 0AA", which is a special valid postcode.
我们想出了这个:
/^([A-PR-UWYZ][A-HK-Y0-9](?:[A-HJKS-UW0-9][ABEHMNPRV-Y0-9]?)?\s*[0-9][ABD-HJLNP-UW-Z]{2}|GIR\s*0AA)$/i
但是注意-这允许组之间有任意数量的空格。
虽然这里有很多答案,但我对其中任何一个都不满意。他们中的大多数只是简单地坏了,太复杂或只是坏了。
我看了@ctwheels的答案,我发现它非常具有解释性和正确性;我们必须为此感谢他。然而,对我来说,如此简单的事情又有太多的“数据”了。
幸运的是,我设法获得了一个数据库,其中仅包含英国的100多万个活动邮政编码,并编写了一个小型PowerShell脚本来测试和基准测试结果。
英国邮政编码规格:有效的邮政编码格式。
这是“我的”正则表达式:
^([a-zA-Z]{1,2}[a-zA-Z\d]{1,2})\s(\d[a-zA-Z]{2})$
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
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)