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

根据皇家邮政的程序员指南，检查邮政编码是否为有效格式:

          |----------------------------outward code------------------------------| |------inward code-----|
#special↓       α1        α2    AAN  AANA      AANN      AN    ANN    ANA (α3)        N         AA
^(GIR 0AA|[A-PR-UWYZ]([A-HK-Y]([0-9][A-Z]?|[1-9][0-9])|[1-9]([0-9]|[A-HJKPSTUW])?) [0-9][ABD-HJLNP-UW-Z]{2})$

uk上的所有邮编都匹配，除了那些不再使用的邮编。

增加一个?空格后，使用不区分大小写的匹配来回答这个问题:

'se50eg'.match(/^(GIR 0AA|[A-PR-UWYZ]([A-HK-Y]([0-9][A-Z]?|[1-9][0-9])|[1-9]([0-9]|[A-HJKPSTUW])?) ?[0-9][ABD-HJLNP-UW-Z]{2})$/ig);
Array [ "se50eg" ]

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;

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

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。

这里的大多数答案都不能适用于我数据库中的所有邮政编码。我终于找到了一个验证与所有，使用政府提供的新正则表达式:

https://www.gov.uk/government/uploads/system/uploads/attachment_data/file/413338/Bulk_Data_Transfer_-_additional_validation_valid_from_March_2015.pdf

在之前的答案中都没有，所以我把它贴在这里，以防他们把链接拿下来:

^([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])))) [0-9][A-Za-z]{2})$

更新:更新的正则表达式由杰米公牛指出。不确定这是我的错误复制或它是一个错误在政府的正则表达式，链接是现在…

更新:正如ctwheels发现的那样，这个正则表达式与javascript的正则表达式兼容。请参阅他的评论，了解一个适用于pcre (php)风格的评论。