Hi,
Using request.route_url() I’m trying to generate a URL of the form:
_app_url/#/foo?arg=bla
Using a route="/#/foo" escaped the hash, and so did "/{hash_}/foo", and
that's according to the change comment
<https://docs.pylonsproject.org/projects/pyramid/en/latest/api/request.html#pyramid.request.Request.route_url>
for Pyramid v1.9. Furthermore, an _anchor is always placed *after* query
string arguments and must not be empty, so using _anchor="" is also not an
option.
The best I can come up with is assembling such a URL like so:
_app_url + "/#/foo?" + pyramid.url.urlencode({"arg": "bla"})
but that looks rather ugly.
What is the recommended way of generating such a URL string? (Leaving aside
the sense or nonsense of such a URL…)
Thanks!
Jens
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