CODEWITHSUNDEEP
pythoninput
spapadop
spapadop

Reputation: 61

How can you bypass an input prompt in Python3 when there is no option to skip the input?

Using a library that when I call a certain function that I will call foo() , I am prompted for input as such:

def foo():
   if input("Proceed?") == 'yes':
        take_action()

I am required to pass in an answer to continue.

But I would like to be able to loop over foo() with a default value to feed the prompt.

The problem is the developers did not provide an option to supply a default value for the prompt.

Ideally they would have written foo() like this:

def foo(default_response=None):
    if default_response == 'yes':
        take_action()
    elif input("Proceed?") == 'yes':
        take_action()

Given that they did not supply an option for a default response, is there a way for me to loop over foo() and provide the input automatically without changing the source code?

Upvotes: 0

Views: 1399

Answers (1)

larsks
larsks

Reputation: 312610

One solution is to monkey patch the library to temporarily replace the input function. For example, if I have module foo.py that looks like this:

def take_action():
    print("doing something")


def foo():
    if input("Proceed? ") == "yes":
        take_action()

I can write bar.py like this:

import foo
from unittest import mock


def fake_input(default_response=None):
    '''Creates a replacement for the `input` function that will
    return a default response if one was provided.'''

    def _input(prompt):
        return default_response if default_response else input(prompt)

    return _input


with mock.patch("foo.input", new=fake_input("yes")):
    foo.foo()

foo.foo()

If you run this code, you'll see that the first call to foo.foo() uses a default input, and proceeds without prompting. The second call to foo.foo() will prompt normally.

Upvotes: 1

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