Regex for fixed width field

天涯浪子 提交于 2019-12-01 06:42:33

Regular expression are not the answer to every single problem. My advice would be to do something like:

boolean isValidSomethingOrOther (string):
    if string.length() != 4:
        return false
    for each character in string:
        if not character.isNumeric():
            return false
    if string.toInt() > 1331:
        return false
    return true

If you must use a regex, there's nothing wrong with your solution but I'd probably use the following variant (just based on my understanding of RE engines and how they work):

^0[0-9]{3}|1[0-2][0-9]{2}|13[0-2][0-9]|133[01]$
  • The first section matches 0000-0999.
  • The second matches 1000-1299.
  • The third matches 1300-1329.
  • The final one matches 1330 and 1331.

Update:

Just on the elegance comment, there are many forms of elegance of which regexes are one. You can also achieve elegance just by abstracting the validation out to a separate function or macro and then call it from your code:

if isValidSomethingOrOther(str) ...

where SomethingOrOther is a concrete business object. This allows you to change your idea of a valid object easily, even using a regex as you desire or any other checks you deem appropriate (such as my function above).

This allows you to cater for any changes down the line such as the requirement that these object now have to be prime numbers.

I'm sure I could write a "prime-number-less-than-1332" regex. I'm equally sure I wouldn't want to - I'd prefer to code that up as a function (or lookup table for raw speed), especially since the regex would most likely just look like:

^2|3|5|7| ... |1327$

anyway.

This seems too easy, am I understanding the problem correctly?

\[01][0-9]{3}\

I don't know what .. means, integer in range? That must be a perlism or something.

This seems to work the way you want to me:

In [3]: r = re.compile(r'[01][0-9]{3}')

In [4]: r.match('0001')
Out[4]: <_sre.SRE_Match object at 0x2fa2d30>

In [5]: r.match('1001')
Out[5]: <_sre.SRE_Match object at 0x2fa2cc8>

In [6]: r.match('2001')

In [7]: r.match('001')

In [8]: 
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!