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
根据维基百科的表格
这种模式适用于所有情况
(?:[A-Za-z]\d ?\d[A-Za-z]{2})|(?:[A-Za-z][A-Za-z\d]\d ?\d[A-Za-z]{2})|(?:[A-Za-z]{2}\d{2} ?\d[A-Za-z]{2})|(?:[A-Za-z]\d[A-Za-z] ?\d[A-Za-z]{2})|(?:[A-Za-z]{2}\d[A-Za-z] ?\d[A-Za-z]{2})
当在Android / Java上使用它时,使用\\d
其他回答
^([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
前半段邮政编码有效格式
[a - z] [a - z][0 - 9]的[a -ž] [a - z] [a - z] [0 - 9] [0 - 9] [a - z] [0 - 9] [0 - 9] [a - z] [a - z] [0 - 9] [a - z] [a - z]的[a -ž] [a - z][0 - 9]的[a -ž] [a - z] [0 - 9]
异常 位置1 - QVX未使用 位置2 -除GIR 0AA外,IJZ不使用 位置3 - AEHMNPRTVXY只使用 位置4 - ABEHMNPRVWXY
邮政编码的后半部分
[0 - 9] [a - z]的[a -ž]
异常 位置2+3 - CIKMOV未使用
记住,不是所有可能的代码都被使用了,所以这个列表是有效代码的必要条件,而不是充分条件。只是匹配所有有效代码的列表可能会更容易?
我从一个XML文档中窃取了这个,它似乎涵盖了没有硬编码的GIRO的所有情况:
%r{[A-Z]{1,2}[0-9R][0-9A-Z]? [0-9][A-Z]{2}}i
(Ruby语法忽略大小写)
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。
我一直在寻找一个英国邮政编码正则表达式的最后一天左右,无意中发现了这个线程。我尝试了上面的大部分建议,但没有一个对我有用,所以我想出了自己的正则表达式,据我所知,它捕获了截至1月13日的所有有效的英国邮政编码(根据皇家邮政的最新文献)。
The regex and some simple postcode checking PHP code is posted below. NOTE:- It allows for lower or uppercase postcodes and the GIR 0AA anomaly but to deal with the, more than likely, presence of a space in the middle of an entered postcode it also makes use of a simple str_replace to remove the space before testing against the regex. Any discrepancies beyond that and the Royal Mail themselves don't even mention them in their literature (see http://www.royalmail.com/sites/default/files/docs/pdf/programmers_guide_edition_7_v5.pdf and start reading from page 17)!
注意:在皇家邮政自己的文献中(链接以上),第3和第4位的位置略有模糊,如果这些字符是字母,则例外。我直接联系了皇家邮政,用他们自己的话说,“AANA NAA格式的出境代码的第4个位置的信件没有例外,而第3个位置的例外只适用于ANA NAA格式的出境代码的最后一个字母。”直接从马嘴里说出来的!
<?php
$postcoderegex = '/^([g][i][r][0][a][a])$|^((([a-pr-uwyz]{1}([0]|[1-9]\d?))|([a-pr-uwyz]{1}[a-hk-y]{1}([0]|[1-9]\d?))|([a-pr-uwyz]{1}[1-9][a-hjkps-uw]{1})|([a-pr-uwyz]{1}[a-hk-y]{1}[1-9][a-z]{1}))(\d[abd-hjlnp-uw-z]{2})?)$/i';
$postcode2check = str_replace(' ','',$postcode2check);
if (preg_match($postcoderegex, $postcode2check)) {
echo "$postcode2check is a valid postcode<br>";
} else {
echo "$postcode2check is not a valid postcode<br>";
}
?>
我希望它能帮助其他遇到这条线索寻找解决方案的人。