Match string after second occurrence of '/'

Question

I'm trying to do the following:

Let's say we have an iterator returning strings looking like this:

/*/*/*/*/*/

where * can be any string. I would like a match if the second * is equal to some arbitrary string, lets say 'test'.

/*/test/*/*/*/  <--- match

Answer 1

If you really need a regex, then you can do it this way:

def check(s):
    return re.match(r"\/[^\/]*\/test\/[^\/]*\/[^\/]*\/[^\/]*\/", s) is None

print(check("/one/test/three/four/five/"))
print(check("/one/two/three/four/five/"))

Output:

False
True

This requires that there is exactly the pattern /*/test/*/*/*/ , where * is everything except for / .

Answer 2

This should do the job

def check(s, index, searched_match):
    return s.split('/')[index] == searched_match

print(check("/*/test/*/*/*/", 2, "test"))
>>> True

You can also use the maxsplit parameter of the split method (use it only if you are sure that there is something after test ):

def check(s, index, searched_match):
    return s.split('/', index + 1)[-2] == searched_match