字符串正则表达式匹配 (DOI)

Question

Hi I'm struggling to understand why my Regex isn't working.

I have URL's that have DOI's on them like so:

https://link.springer.com/10.1007/s00737-021-01116-5
https://journals.sagepub.com/doi/pdf/10.1177/1078390319877228
https://journals.sagepub.com/doi/pdf/10.1177/1078390319877228
https://onlinelibrary.wiley.com/doi/10.1111/jocn.13435
https://journals.sagepub.com/doi/pdf/10.1177/1062860613484171
https://onlinelibrary.wiley.com/resolve/openurl?genre=article&title=Natural+Resources+Forum&issn=0165-0203&volume=26&date=2002&issue=1&spage=3
https://dx.doi.org/10.1108/14664100110397304?nols=y
https://onlinelibrary.wiley.com/doi/10.1111/jocn.15833
https://www.tandfonline.com/doi/pdf/10.1080/03768350802090592?needAccess=true

And I'm using for example this Regex, but it always returns empty?

print(re.findall(r'/^10.\d{4,9}/[-._;()/:A-Z0-9]+$/i', 'https://dx.doi.org/10.1108/02652320410549638?nols=y'))

Where have I gone wrong?

Answer 1

It looks like you come from another programming language that has the notion of regex literals that are delimited with forward slashes and have the modifiers following the closing slash (hence /i ).

In Python there is no such thing, and these slashes and modifier(s) are taken as literal characters. For flags like i you can use the optional flags parameter of findall .

Secondly, ^ will match the start of the input string, but evidently the URLs you have as input do not start with 10 , so that has to go. Instead you could require that the 10 must follow a word break... ie it should not be preceded by an alphanumerical character (or underscore).

Similarly, $ will match the end of the input string, but you have URLs that continue with URL parameters, like ?nols=y , so again the part you are interested in does not go on until the end of the input. So that has to go too.

The dot has a special meaning in regex, but you clearly intended to match a literal dot, so it should be escaped.

Finally, alphanumerical characters can be matched with \w , which also matches both lower case and capital Latin letters, so you can shorten the character class a bit and do without any flags such as i ( re.I ).

This leaves us with:

print(re.findall(r'\b10\.\d{4,9}/[-.;()/:\w]+', 
                'https://dx.doi.org/10.1108/02652320410549638?nols=y'))