get the name of the argument that was sent

Asked Jan 06 '23 at 15:53

Active Jan 06 '23 at 16:39

Viewed 24 times

I need a way to get the parameter that was sent name.
for example

class Message:  
    I = 1
    LIKE = 2
    COODING = 3

sendData(Message.LIKE)


def sendData(arg):
    print(arg.name) -> Message.LIKE
    print(arg.value) -> 2

And before some says use class Message(Enum): this is prebuilt Class so I can't change the class itself.

I have tried stuff like

sendData(enum(Message.LIKE))

and check via inspect module the method call signature with no luck

EDIT: This is no duplicated, as I says. I cant change class Message to class Message(Enum)

edited Jan 06 '23 at 16:39

asked Jan 06 '23 at 15:53

Bar Benezri

If you want strings, why not just use those as the values, rather than numbers? – Barmar Jan 06 '23 at 15:56
cus i need the number and the string – Bar Benezri Jan 06 '23 at 15:57
`Message.LIKE` is not a name; it's an expression used to generate a value of type `int`. The value `2` bound to `arg` knows nothing about how it was provided to `sendData`. – chepner Jan 06 '23 at 16:01
(For all `sendData` knows, `Message` wasn't even involved: it could have been called like `sendData(2)`.) – chepner Jan 06 '23 at 16:01
I already know this. I'm looking for way to make Message.LIKE to Enum single instance – Bar Benezri Jan 06 '23 at 16:40

get the name of the argument that was sent

0 Answers0