Python string templater

Question

I'm using this REST web service, which returns various templated strings as urls, for example:

"http://api.app.com/{foo}"

In Ruby, I can then use

url = Addressable::Template.new("http://api.app.com/{foo}").expand('foo' => 'bar')

to get

"http://api.app.com/bar"

Is there any way to do this in Python? I know about %() templates, but obviously they're not working here.

Answer 1

In python 2.6 you can do this if you need exactly that syntax

from string import Formatter
f = Formatter()
f.format("http://api.app.com/{foo}", foo="bar")

If you need to use an earlier python version then you can either copy the 2.6 formatter class or hand roll a parser/regex to do it.

Answer 2

Don't use a quick hack.

What is used there (and implemented by Addressable) are URI Templates. There seem to be several libs for this in python, for example: uri-templates. described_routes_py also has a parser for them.