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
看看本页的python代码:
http://www.brunningonline.net/simon/blog/archives/001292.html
I've got some postcode parsing to do. The requirement is pretty simple; I have to parse a postcode into an outcode and (optional) incode. The good new is that I don't have to perform any validation - I just have to chop up what I've been provided with in a vaguely intelligent manner. I can't assume much about my import in terms of formatting, i.e. case and embedded spaces. But this isn't the bad news; the bad news is that I have to do it all in RPG. :-( Nevertheless, I threw a little Python function together to clarify my thinking.
我用它来处理邮政编码。
以下是我们处理英国邮政编码问题的方法:
^([A-Za-z]{1,2}[0-9]{1,2}[A-Za-z]?[ ]?)([0-9]{1}[A-Za-z]{2})$
解释:
期望有1或2个a-z字符,上或下都没问题 预期有1到2个数字 期望0或1个a-z字符,上或下精细 允许使用可选空间 期望1个数字 期望有2个a-z,上下都没问题
这将获得大多数格式,然后我们使用db来验证邮政编码是否真实,该数据由openpoint https://www.ordnancesurvey.co.uk/opendatadownload/products.html驱动
希望这能有所帮助
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
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。
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