Python calling function inside other function not working

Question

This is what I have right now:

def open_file():
    file_name = input("What is the name of the file you want to open?")
    while True:
        try:
            file = open(file_name, 'r')
            header = file.readline()
            return (file)
            break
        except FileNotFoundError:
            file_name = input("What is the name of the file you want to open?")

def process_file():
    file = open_file()
    print(file)

def main():
    process_file()

I can't even get it to get to prompting me for the file name I want to open. Doesn't that mean it's not a problem with my loop but the way I am calling my functions?

Unlike C, `main()` isn't called automatically. – John Gordon Feb 29 '16 at 20:28 — John Gordon, Feb 29 '16 at 20:28

score 3 · Answer 1 · edited May 23 '17 at 10:28

3

Assuming you called your module something like foo.py

You could add this bit of code:

if __name__ == '__main__':
    main()

And then in the console run:

  >>python foo.py

For an explanation of why this works, you can see this answer: What does if __name__ == "__main__": do?

edited May 23 '17 at 10:28

Community

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answered Feb 29 '16 at 21:24

Juan

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score 1 · Accepted Answer · answered Feb 29 '16 at 20:28

1

You need to call main() to run your code. If you do that it works just fine.

answered Feb 29 '16 at 20:28

Mark Skelton

3,663
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Leo · Answer 3 · 2016-02-29T20:46:15.973

1

You need to call main() in order for your code to run.

edited Feb 29 '16 at 20:46

answered Feb 29 '16 at 20:34

Leo

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raw_input doesn't exist in Python 3 – Feb 29 '16 at 20:39

Python calling function inside other function not working

3 Answers3