Python - Flask Default Route possible?

Question

In Cherrypy it\'s possible to do this:

@cherrypy.expose
def default(self, url, *suburl, **kwarg):
    pass

Is there a flask equivalent?

Answer 1

There is a snippet on Flask's website about a 'catch-all' route for flask. You can find it here.

Basically the decorator works by chaining two URL filters. The example on the page is:

@app.route('/', defaults={'path': ''})
@app.route('/')
def catch_all(path):
    return 'You want path: %s' % path

Which would give you:

% curl 127.0.0.1:5000          # Matches the first rule
You want path:  
% curl 127.0.0.1:5000/foo/bar  # Matches the second rule
You want path: foo/bar