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"}

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。

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*$)

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

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。

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。