Returning value from a function when the function itself called inside PYTHON

Question

My function in python looks like below:

def choose_sepa_reason_code():
    # This method is for providing proper sepa reason code

    sepa_reason_code = input("\nPlease provide SEPA reason code to reject the payment: ")

    if sepa_reason_code in sepa_reason_codes:
        return sepa_reason_code
    else:
        print("Your reason codes doesnt match the list:\n")
        pprint(sepa_reason_codes)
        choose_sepa_reason_code()

provided_sepa_reason_code = choose_sepa_reason_code()

The if statement is to make sure that user will provide proper code, but if he is wrong at the first time my function later returns None.

Could you please advise how to change it to get final correct user input?

first you don't return anything from your else branch, second you shouldn't call your function recursively. Rewriting the routine with a loop would solve that. — Jean-François Fabre
– Jean-François Fabre ♦, Commented Feb 7, 2017 at 13:48
See Asking the user for input until they give a valid response — roganjosh
– roganjosh, Commented Feb 7, 2017 at 13:49
Possible duplicate of Asking the user for input until they give a valid response — Łukasz Rogalski
– Łukasz Rogalski, Commented Feb 7, 2017 at 13:55
your indentation is broken. Is all the code supposed to be part of choose-sepa_reason_code? — Bryan Oakley
– Bryan Oakley, Commented Feb 7, 2017 at 14:23

Danix · Accepted Answer · 2017-02-07 14:20:10Z

1

If you really want to call the function recursively, call choose_sepa_reason_code() as return value in your else statement:

sepa_reason_code = input("\nPlease provide SEPA reason code to reject the payment: ")

if sepa_reason_code in sepa_reason_codes:
    return sepa_reason_code
else:
    print("Your reason codes doesnt match the list:\n")
    print(sepa_reason_codes)
    return choose_sepa_reason_code()

edited Feb 7, 2017 at 14:20

answered Feb 7, 2017 at 14:07

Danix

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3 Comments

WhatsThePoint Over a year ago

wasnt the whole point in him recalling the function because he wanted it to run until the user enters correctly?

roganjosh Over a year ago

@WhatsThePoint which is what this does.

Danix Over a year ago

@WhatsThePoint Indeed, I would also prefer a solution with a loop, but the OP question was about a recursive one.

WhatsThePoint · Accepted Answer · 2017-02-07 14:00:34Z

1

firstly your code wont run because it isnt indented, secondly you cant call a function inside itself.

this would be a better approach to your task:

def choose_sepa_reason_code():
# This method is for providing proper sepa reason code
    while True: #will provide permanent loop
        sepa_reason_code = input("\nPlease provide SEPA reason code to reject the payment: ")

        if sepa_reason_code in sepa_reason_codes:
            return sepa_reason_code #stops loop and returns the reason code
        else:
            print("Your reason codes doesnt match the list:\n")
            print(sepa_reason_codes)

edited Feb 7, 2017 at 14:00

answered Feb 7, 2017 at 13:55

WhatsThePoint

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9 Comments

WhatsThePoint Over a year ago

thanks, added the while loop and forgot to indent his code :)

WhatsThePoint Over a year ago

also side note to the OP your names of variables, function etc are too similar just with minor differences, makes it harder to read, i would recommend changing them

roganjosh Over a year ago

No prob. A point of note though. You write "secondly you cant call a function inside itself." but this is not true. See this link about recursion

WhatsThePoint Over a year ago

yeah i know technically you can but in most cases you shouldnt because it just starts itself again executes and finishes the original, which isnt a very good approach

roganjosh Over a year ago

"Can't" and "but in most cases you shouldn't" are different things :) The title of the post is quite generic for someone who might be Googling the subject and they may well come across your answer which states it "can't be done".

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