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
接受的答案反映了皇家邮政给出的规则，尽管正则表达式中有一个拼写错误。这个错字似乎在gov.uk网站上也有(就像在XML存档页面中一样)。
在格式A9A 9AA中，规则允许在第三个位置出现P字符，而正则表达式不允许这样。正确的正则表达式应该是:
(GIR 0AA)|((([A-Z-[QVX]][0-9][0-9]?)|(([A-Z-[QVX]][A-Z-[IJZ]][0-9][0-9]?)|(([A-Z-[QVX]][0-9][A-HJKPSTUW])|([A-Z-[QVX]][A-Z-[IJZ]][0-9][ABEHMNPRVWXY])))) [0-9][A-Z-[CIKMOV]]{2})
将其缩短为以下正则表达式(使用Perl/Ruby语法):
(GIR 0AA)|([A-PR-UWYZ](([0-9]([0-9A-HJKPSTUW])?)|([A-HK-Y][0-9]([0-9ABEHMNPRVWXY])?))\s?[0-9][ABD-HJLNP-UW-Z]{2})
它还在第一个和第二个块之间包含一个可选的空格。
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。
我看了一下上面的一些答案，我不建议使用@Dan的答案(12月15日至10日)中的模式，因为它错误地将近0.4%的有效邮政编码标记为无效，而其他的则没有。
军械测量所提供的代号开放服务包括:
包含英国所有当前邮政编码单位的列表
我使用grep从这些数据中针对完整的邮政编码列表(7月6日至13日)运行了上面的每个正则表达式:
cat CSV/*.csv |
# Strip leading quotes
sed -e 's/^"//g' |
# Strip trailing quote and everything after it
sed -e 's/".*//g' |
# Strip any spaces
sed -E -e 's/ +//g' |
# Find any lines that do not match the expression
grep --invert-match --perl-regexp "$pattern"
邮政编码共有1,686,202个。
以下是与每个$模式不匹配的有效邮政编码的数量:
'^([A-PR-UWYZ0-9][A-HK-Y0-9][AEHMNPRTVXY0-9]?[ABEHMNPRVWXY0-9]?[0-9][ABD-HJLN-UW-Z]{2}|GIR 0AA)$'
# => 6016 (0.36%)
'^(GIR ?0AA|[A-PR-UWYZ]([0-9]{1,2}|([A-HK-Y][0-9]([0-9ABEHMNPRV-Y])?)|[0-9][A-HJKPS-UW]) ?[0-9][ABD-HJLNP-UW-Z]{2})$'
# => 0
'^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}$'
# => 0
当然，这些结果只处理被错误地标记为无效的有效邮政编码。所以:
'^.*$'
# => 0
在过滤无效邮编方面，我并没有说哪种模式是最好的。
我需要一个可以在SAS中使用PRXMATCH和相关函数的版本，所以我想到了这个:
^[A-PR-UWYZ](([A-HK-Y]?\d\d?)|(\d[A-HJKPSTUW])|([A-HK-Y]\d[ABEHMNPRV-Y]))\s?\d[ABD-HJLNP-UW-Z]{2}$
测试用例和注意事项:
/*
Notes
The letters QVX are not used in the 1st position.
The letters IJZ are not used in the second position.
The only letters to appear in the third position are ABCDEFGHJKPSTUW when the structure starts with A9A.
The only letters to appear in the fourth position are ABEHMNPRVWXY when the structure starts with AA9A.
The final two letters do not use the letters CIKMOV, so as not to resemble digits or each other when hand-written.
*/
/*
Bits and pieces
1st position (any): [A-PR-UWYZ]
2nd position (if letter): [A-HK-Y]
3rd position (A1A format): [A-HJKPSTUW]
4th position (AA1A format): [ABEHMNPRV-Y]
Last 2 positions: [ABD-HJLNP-UW-Z]
*/
data example;
infile cards truncover;
input valid 1. postcode &$10. Notes &$100.;
flag = prxmatch('/^[A-PR-UWYZ](([A-HK-Y]?\d\d?)|(\d[A-HJKPSTUW])|([A-HK-Y]\d[ABEHMNPRV-Y]))\s?\d[ABD-HJLNP-UW-Z]{2}$/',strip(postcode));
cards;
1 EC1A 1BB Special case 1
1 W1A 0AX Special case 2
1 M1 1AE Standard format
1 B33 8TH Standard format
1 CR2 6XH Standard format
1 DN55 1PT Standard format
0 QN55 1PT Bad letter in 1st position
0 DI55 1PT Bad letter in 2nd position
0 W1Z 0AX Bad letter in 3rd position
0 EC1Z 1BB Bad letter in 4th position
0 DN55 1CT Bad letter in 2nd group
0 A11A 1AA Invalid digits in 1st group
0 AA11A 1AA 1st group too long
0 AA11 1AAA 2nd group too long
0 AA11 1AAA 2nd group too long
0 AAA 1AA No digit in 1st group
0 AA 1AA No digit in 1st group
0 A 1AA No digit in 1st group
0 1A 1AA Missing letter in 1st group
0 1 1AA Missing letter in 1st group
0 11 1AA Missing letter in 1st group
0 AA1 1A Missing letter in 2nd group
0 AA1 1 Missing letter in 2nd group
;
run;
下面的方法将检查邮政编码并提供完整的信息
const isValidUKPostcode = postcode => {
try {
postcode = postcode.replace(/\s/g, "");
const fromat = postcode
.toUpperCase()
.match(/^([A-Z]{1,2}\d{1,2}[A-Z]?)\s*(\d[A-Z]{2})$/);
const finalValue = `${fromat[1]} ${fromat[2]}`;
const regex = /^([Gg][Ii][Rr] 0[Aa]{2})|((([A-Za-z][0-9]{1,2})|(([A-Za-z][A-Ha-hJ-Yj-y][0-9]{1,2})|(([AZa-z][0-9][A-Za-z])|([A-Za-z][A-Ha-hJ-Yj-y][0-9]?[A-Za-z]))))[0-9][A-Za-z]{2})$/i;
return {
isValid: regex.test(postcode),
formatedPostCode: finalValue,
error: false,
message: 'It is a valid postcode'
};
} catch (error) {
return { error: true , message: 'Invalid postcode'};
}
};
console.log(isValidUKPostcode('GU348RR'))
{isValid: true, formattedPostcode: "GU34 8RR", error: false, message: "It is a valid postcode"}
console.log(isValidUKPostcode('sdasd4746asd'))
{error: true, message: "Invalid postcode!"}
valid_postcode('787898523')
result => {error: true, message: "Invalid postcode"}
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