Twig - Get URL for canonical tag

Question

I'm looking to create a dynamic rel="canonical" tag in my application which pulls in the current URL but want to ensure any query parameters are removed. E.g http://www.example.com/test/?page=2 should have the canonical as http://www.example.com/test/, therefore {{ app.request.uri }} doesn't work as this pulls in ?page=2 as well.

Does anyone know how to pull in the absolute path of a page without the query parameters?

Hypothec

I don't think it's possible (maybe I'm wrong), but you could write Twig extension that splits app.request.uri by ? char... – Ace

I've just remembered that there is app.request.baseUrl. Have you tried that? – Ace

Answer 1

This will work,

{{ url(app.request.attributes.get('_route'), app.request.attributes.get('_route_params')) }}

Dias

+1 Haven't tried it but this seems to be really neat solution :) – Ace

Aren't query strings provided in the route params? – Lecce

@SteveBuzonas, nope it's only generate url base on current route name and it's params. If u have additional query strings u need to add it by yourself – Dias

In your 404 page if you extend the template this code is used in, make sure to put this inside an if block or 404 page will become a 500 page because route doesn't exist. – Osbourne

it's ugly as sin, but it still gets the job done in Symfony 4.4. Thanks! – Birdman

Answer 2

I just tried to dump baseUrl and can confirm that it does not work.

However, this works:

{{ app.request.getSchemeAndHttpHost ~ app.request.baseUrl ~ app.request.pathInfo }}

I know, it's not pretty but it does the job :)

Ace

This is what i am looking for, above solution not including parameters to generated url. – Reformed

this should be the accepted answer, cause the other one will result in 500 errors for every single error page like 403, 404, and so on! – Pilferage