pythonflaskcherrypy

Python - Flask Default Route possible?

In Cherrypy it's possible to do this:

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

Is there a flask equivalent?

Solution

  • 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('/<path:path>')
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