Python regular expressions and greedy matching at the end

Question

I have this simple regular expression substitution just to ensure that some URL ends in a slash character:

   url = re.sub("/*$", "/", "foo/")

...but it just happens that when I run that code, the result is unexpectedly foo//.

After further experimentation the explanation I have found is that the regular expression /*$ matches the slash at the end of the string and replaces it by another slash, but then it matches again the empty string just after the replaced slash.

Is there any simple way to workaround this issue?

Update: Well, It seems you can tell sub how many times you want the replacement done at most:

   url = re.sub("/*$", "/", "foo/", 1)

why do you use re.sub to check a condition? use re.match instead — Leonardo Scotti, Feb 24 '21 at 12:54
If you want to match the last character *not* being a slash, `"[^/]$"`, but why use regex instead of just `str.endswith`? — jonrsharpe, Feb 24 '21 at 12:56
Using `re` is overkill here. Just use: `'foo/'.rstrip('/') + '/'` — James, Feb 24 '21 at 12:57
you have said in the question that you are ensuring, so maybe it will be better to check if the condition is already satisfied and olnly after finding an unmatched requirement start to replace — Leonardo Scotti, Feb 24 '21 at 12:58
As you said yourself, an empty string is matched at the end. Simply change to `/+$`... There is no point in using `*` because you want to replace ***existing*** slashes... — Tomerikoo, Feb 24 '21 at 12:59
@Tomerikoo, that doesn't work when there are no slashes at the end of the input string — salva, Feb 24 '21 at 13:01
In that case James' suggestion with `rstrip` seems to be the one to go with... — Tomerikoo, Feb 24 '21 at 13:06

score 0 · Answer 1 · answered Feb 24 '21 at 13:35

For the sake of a regex answer:

The /*$ matches any number of slashes / at the end of the input, but since the * allows zero repetitions, it also matches the position after the last character in the string, be that a slash or not.

This allows you to append a new slash at the end in the first place, but it will also not prevent adding another slash when there already is one.

And that's the key: You must check that the last character in the string is not a slash. That can be done with a negative look-behind:

(?<!/)/*$

This will replace any number of slashes at the end of the string with a single slash, but without the unnecessary duplication.

...but for all practical purposes, use .rstrip('/') + '/' as @James has suggested in the comments.

score 0 · Answer 2 · answered Feb 24 '21 at 13:36

If you absolutely do want/need to do it with a regex, you need to match the end of the url with a non-slash (one or more). And then replace that with a slash (and the original ending). So will also need a capture group to get that ending:

re.sub("([^/]+)/*$", r"\1/", "foo")
# 'foo/'
re.sub("([^/]+)/*$", r"\1/", "foo/")
# 'foo/'
re.sub("([^/]+)/*$", r"\1/", "foo//")
# 'foo/'
re.sub("([^/]+)/*$", r"\1/", "foo///////////")
# 'foo/'
re.sub("([^/]+)/*$", r"\1/", "bar/foo//")
# 'bar/foo/'

As mentioned by @James in the comments, you're better off using an easier approach:

'foo/'.rstrip('/') + '/'

which strips off trailing slashes (if there are any) and then adds a slash.

Python regular expressions and greedy matching at the end

2 Answers2