I am trying to write a regex that matches a url of the following format:
/api/v1/users/<mongo_object_id>/submissions
Where an example of a mongo_object_id is 556b352f87d4693546d31185.
I have cooked up the following pattern, but it does not seems to work.
/api/v1/users\\/(?=[a-f\\d]{24}$)(\\d+[a-f]|[a-f]+\\d)\\/submissions
Any help is appreciated.
This will do (considering 24 hex chars), using raw keyword before string so no need to escape with double slashes:
r'\/api\/v1\/users\/([a-f\d]{24})\/submissions'
Python console:
>>> re.findall(r'\/api\/v1\/users\/([a-f\d]{24})\/submissions','/api/v1/users/556b352f87d4693546d31185/submissions')
['556b352f87d4693546d31185']
It looks like an object's ID is a hexadecimal number, which means that it's matched by something as simple as this:
[0-9a-f]+
If you want to make sure it's always 24 characters:
[0-9a-f]{24}
Toss that between the slashes:
/api/v1/users/([0-9a-f]{24})/submissions
And it should work.
Note: You will probably have to escape the slashes, depending on how Python's regex syntax works. If I remember right, you can do this:
import re
re.findall(r'/api/v1/users/([0-9a-f]{24})/submissions', url)
or
re.findall(r'/api/v1/users/([0-9a-f]{24})/submissions', url, re.I)
if you wanna make the whole thing case-insensitive.
mongo_object_id?