CODEWITHSUNDEEP
pythonlambdaclosuresbottlegetattr
Basj
Basj

Reputation: 46341

How to have a lambda function evaluate a variable now (and not postponed)

I have a class for a hardware object (here Fridge), and I'd like to automatically create HTTP API routes for a given subset of Fridge's methods, here open and close.

from bottle import Bottle, request

class Fridge:
    def _not_exposed(self):
        print("This one is never called via HTTP")
    def open(self, param1=None, param2=None):
        print("Open", param1, param2)
    def close(self):
        print("Close")
    # + many other methods

f = Fridge()
app = Bottle("")

for action in ["open", "close"]:
    app.route(f"/action/{action}", callback=lambda: (getattr(f, action)(**request.query)))

app.run()

It works, except that in

...callback=lambda: (getattr(f, action)(**request.query))

action is evaluated when the lambda function is called.

Thus when opening http://127.0.0.1:8080/action/open?param1=123, the lambda is called, and at this time action has the value ... "close" (the last value in the for enumeration), then getattr(f, action) links to f.close instead of f.open. This is not what we want!

Question: in this lambda definition, how to have action evaluated now, and not postponed?

Expected behaviour for the for loop:

app.route("/action/open", callback=lambda: f.open(**request.query)))
app.route("/action/close", callback=lambda: f.close(**request.query)))

Upvotes: 3

Views: 46

Answers (0)

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