一个老帖子，但在谷歌的结果仍然很高，所以我认为我应该更新。10月14日的文档将英国邮政编码正则表达式定义为:

^([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})|(([**AZ**a-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})$

来自:

https://www.gov.uk/government/uploads/system/uploads/attachment_data/file/359448/4__Bulk_Data_Transfer_-_additional_validation_valid.pdf

该文档还解释了它背后的逻辑。然而，它有一个错误(粗体)，也允许小写，虽然合法的是不常见的，所以修改版本:

^(GIR 0AA)|((([A-Z][0-9]{1,2})|(([A-Z][A-HJ-Y][0-9]{1,2})|(([A-Z][0-9][A-Z])|([A-Z][A-HJ-Y][0-9]?[A-Z])))) [0-9][A-Z]{2})$

这适用于新的伦敦邮政编码(例如W1D 5LH)，以前的版本没有。

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。

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

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})$

这个允许两边有空格和制表符，以防你不想验证失败，然后在另一边修剪它。

^\s*(([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])))) {0,1}[0-9][A-Za-z]{2})\s*$)

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