Add method to easily generate a pre-filtered URL

[samuelhwilliams]

I don't think it's trivial currently to generate a URL for an admin page with some filters already applied.

I'd like to be able to do something like:

filtered_url = UserModelView.url_for(search="sam", filters={"age": [IntGreaterFilter(30)]})

[ElLorans]

I think this is feasible right now, more easily using https://flask-admin.readthedocs.io/en/stable/api/mod_contrib_sqla/#flask_admin.contrib.sqla.ModelView.named_filter_urls .

[samuelhwilliams]

Yep that definitely helps, and I'm using that in a hacky workaround at the moment, but something more official would be good.

[samialfattani]

I have spent 3 days trying to follow and understand the code and it was really confusing. I believe the part of filtering need a complete code enhancement. many tuples should be converted into objects & classes so that it can be more readalble and the tuple's parts can have meaningfule names.

[samialfattani]

filtered_url = UserModelView.url_for(search="sam", filters={"age": [IntGreaterFilter(30)]})

I still don't understand why you are expecting the filters parameter should have this kind of structure dict[str, list[BaseFilter]]!! Why the filter of IntGreaterFilter(30) is placed inside a list ? and why dictionary ?

What I'm expecting the code is simpler like:

filtered_url = UserModelView.url_for(search="sam", filters=[IntGreaterFilter('age', 30), FilterLike('email', '@example.com') ])

output:
/admin/user/?search=sam&flt0_7=30&flt1_14=example.com

knowing that all provided filters will be connected by ANDs

[samuelhwilliams]

Yep, that's fine, I didn't really sit and think the API through yet. My first comment was mostly just a note about the kind of functionality I want, not to commit to the exact 'how'.