Current approach to work with trailing slashes

[hoefling]

Hi, what is the best (i.e. intended) approach to add and drop trailing slashes? I am currently using

>>> u = yarl.URL('https://www.example.com/spam/eggs')
>>> u
URL('https://www.example.com/spam/eggs')
>>> u / ''
URL('https://www.example.com/spam/eggs/')

for adding and

>>> u = yarl.URL('https://www.example.com/spam/eggs/')
>>> u
URL('https://www.example.com/spam/eggs/')
>>> u.with_path(u.path.removesuffix('/'))
URL('https://www.example.com/spam/eggs')

for dropping the slashes, but maybe somebody has a more elegant solution, esp. for the dropping part.

[webknjaz]

@commonism might have opinions on this ^

[commonism]

for adding I don't think it'll get any better than

u / ''

which is even safe, as it does not add a / if the path ends with / already
but could be improved to avoid unnecessary construction

u / '' if u.parts[-1] != "" else u

removing / - possible via

(u / "").parent

or - avoiding the cost of construction if not required:

u.parent if u.parts[-1] == "" else u

but it takes a certain angle to recognize elegance in this I guess

[hoefling]

@commonism uh I like the (u / "").parent one 👍

[hoefling]

@commonism disregard that, (u / "").parent does not remove trailing slash from host.

>>> u = yarl.URL('https://www.example.com/')
>>> u.parent
URL('https://www.example.com/')

[commonism]

This is partial due to the requirements of RFC3986 - 6.2.3 - Scheme-Based Normalization

   In general, a URI that uses the generic syntax for authority with an
   empty path should be normalized to a path of "/".

Currently this is incomplete in yarl as

>>> yarl.URL("http://www.example.org")
URL('http://www.example.org')

does not add it.

Therefore currently

(u / '').parent if u.path != "/" else u.with_path("")

will do, but it will break once scheme based normalization is complete in yarl.

On the other hand - there is no URL without trailing / in http, maybe completing scheme-based normalization will do as well?