No way to replace path without replacing query

[Lucretiel]

The following is unexpected behavior:

url1 = URL("http://www.example.com/foo?bar=10")
url2 = url1.with_path("/bar")

assert url1.query = url2.query  # Assertion fails

While I understand the design decision to treat the query and fragment components as part of the path, this has left us without an (efficient) way to preserve the query and fragment components when replacing just the path. The only way to do it is:

url2 = url1.with_path(new_path).with_query(url1.query_string).with_fragment(url1.fragment)

This is a total of 4 extra method calls, at least of which does a lot of unnecessary processing. I'm assuming that there's code out there that depends on this weird behavior, so I'd like to propose and, with permission, implement a new method, with_just_path, which replaces the path but leaves the fragment and query the same.

[asvetlov]

Did you try URL.build?

[Lucretiel]

I mean, that would work, but why even have the with_* methods in the first place? This seems like it'd fit in well with the other with methods. It's even MORE verbose, because I have to list out every single field in the URL in the build call.

[asvetlov]

Honestly I don't want to blow up URL class with new methods.
Also I suspect your need is very rare: usually query/fragment is described by path, different paths have different allowed query params.
The only exception is passing in query some constant information like security token, it could be done by url.update_query() now.

[Lucretiel]

My need is in use for a web framework, which is why I'm interested in reducing the amount of unnecessary work implied by with_query. In particular, I'm working on django-style path routing where the front of a path is cut off when it's passed to child route views. It's not exactly a very common use case, but I can imagine it'd be common for creating web frameworks based on yarl.

[asvetlov]

Thus do you need cutting first path symbols from relative URL?
Make sense.
Maybe create a special method for this case?

[eternal-sorrow]

Then URL.with_host() should cut off the path, query and fragment altogether. Why the inconsistency? It doesn't make sense.

>>> URL('https://foo.tld/path?a=1#fragment').with_host('bar.tld')
URL('https://bar.tld/path?a=1#fragment')
>>> URL('https://foo.tld/path?a=1#fragment').with_path('/new/path')
URL('https://foo.tld/new/path')

[apalala]

It is totally unexpected that .with_path() will remove query.

[JP-Ellis]

Working with this library, and got this unexpected behaviour as well. The docs quite clearly say that with_query should:

Return a new URL with path replaced, encode path if needed.

Is there any appetite for a PR?

[trim21]

is this considered as a bug or by-design？

[asvetlov]

It is a design decision.
I hear the need for path replacement with saving the query though.
We cannot just change the behavior, it will break existing code. What we can do is change URL.with_path() method to URL.with_path(path, *, keep_query=False).
Passing keep_query=True saves an existing query string.
Does it sound good?

[trim21]

sounds good，maybe also fragment …

[paul-nameless]

Hi, I made some changes to support this behavior. Could you please take a look? Thank you!
#1421