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 -- You received this message because you are subscribed to the Google Groups "pylons-discuss" group. To unsubscribe from this group and stop receiving emails from it, send an email to pylons-discuss+unsubscr...@googlegroups.com. To post to this group, send email to pylons-discuss@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/pylons-discuss/09785b15-32f1-4f9a-98f5-67b77caa0080%40googlegroups.com. For more options, visit https://groups.google.com/d/optout.