Expected: http://some_url.com/api/v1/1/2/3/4" -> http://some_url.com/api/v1/*/*/*/*/
What I use:
re.sub(r"/\d+/?", "/*/", str(url), flags=re.IGNORECASE)
Actual: http://some_url.com/api/v1/*/2/*/4/
regex: eliminate all numbers between slashes
Viewed 60
2 Answers
You may use
/\d+(?=/|$)
/\d+(?![^/])
and replace with /*. See the regex demo.
In Python:
url = re.sub(r"/\d+(?=/|$)", "/*", url)
Details
/- a/char\d+- 1+ digits(?![^/])- a negative lookahead that fails the match if the next char is not a character other than/(so, end of string or/are required immediately to the right of the current location, same as with the positive(?=/|$)lookahead)
See Python demo online:
import re
url = 'http://some_url.com/api/v1/1/2/3/4'
url = re.sub(r"/\d+(?=/|$)", "/*", url)
print(url)
# => http://some_url.com/api/v1/*/*/*/*
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