pyramid: constructe route_url without request object

Question

I use yapps to generate a parser for a LaTex-ish language (for example to translate stuff like \begin{itemize} to the corresponding <ul>-Tags) within pyramid. One command (i.e. \ref{SOMEID}) should construct a route via a call of route_url (or route_path) and pass the id to it. Since this call happens deep in the code that was generated by yapps and the grammar that I defined, I don't see any possibility to pass a request object to it.

Is there some sort of global request object? Or, since I foresee that I shouldn't use it, is there a possibility to construct a route (that depends on a parameter) without a request object?

Trolly

Answer 1

route_url requires both a request and a registry (request.registry). It generates urls relative to the request, and it accesses the list of all routes and other settings from the registry. Thus, you must generate a dummy request with parameters you care about. For example:

from pyramid.request import Request

request = Request.blank('/', base_url='https://example.com/prefix')
request.registry = config.registry

Now you can store this request anywhere, it's good to go representing everything about your site: the hostname/port (example.com:443), the prefix your app is mounted at (/prefix), the uri scheme (https).

If you need to get this deep down into your code you may have to make it a global or attach it to some context/registry that you have available, but what I've shown is how to make the request that you require.

Donavon

thank you very much, I didn't know the blank-function. I used get_current_registry() from pyramid.threadlocal to get the registry and then could just call route_path. – Trolly