How to return status code in Python without actually exiting the process?

Question

I'm trying to return a status code without exiting the process of my script. Is there an equivalent to sys.exit() that does not stop execution ?

What I'd like to happen :

If no exception is raised, status code is by default 0. For minor exceptions, I want to return status code 1, but keep the process running , unless a more critical error exits the process and returns status code 2 with sys.exit(2) . In my case, I'm looping through a bunch of PDF files and if one file has not been anonymised for some reason, I want to return status code 1 while looping through the rest of the files.

In other words, status code should correspond to :

• 0 - SUCCESS : all files have been sucessfully anonymised
• 1 - WARNING : at least one file has not been anonymised
• 2 - ERROR : critical error, interrupted process

For clarity : by status code I mean, what is returned by the Bash command echo $? after the execution of the script.

What I have done : I have tried using sys.exit(1) to handle minor exceptions but it automatically stops execution, as stated in the library documentation.

To sum up with code snippets :

For a given main.py :

import sys
import os

def anonymiser(file) -> Boolean :
    # Returns True if successfully anonymises the file, False otherwise
    ...

def process_pdf_dir():
    for file in os.listdir('pdf_files/'):
        if file.endswith('.pdf'):
            if anonymiser(file) == False:
                '''
                Here goes the code to change status code to 1 
                but keeps looping through the rest of the files
                unlike `sys.exit(1)` that would stop the execution.
                '''
            if criticalError = True:
                sys.exit(2)             

if __name__ == '__main__':
    process_pdf_dir()

After calling the script on the terminal, and echoing its status like so :

$ python main.py
$ echo $?

If indeed, at least one file has not been anonymised :

• Expected output (if no critical error) :

   1

• Expected output (if critical error) :

   2

• Current output (if no critical error) :

   0

• Current output (if critical error) :

   2

Hope this is clear enough. Any help would be greatly appreciated. Thanks.

Answer 1

Wait until the end of your loop to return the 1 exit code. I'd do it like:

def process_pdf_dir() -> int:
    ret = 0
    for file in os.listdir('pdf_files/'):
        if file.endswith('.pdf'):

            if not anonymiser(file):
                # remember to return a warning
                ret = 1

            if criticalError:
                # immediately return a failure
                return 2
    return ret

if __name__ == '__main__':
    sys.exit(process_pdf_dir())

If "critical error" includes uncaught exceptions, maybe put that outside of process_pdf_dir , eg:

if __name__ == '__main__':
    try:
        sys.exit(process_pdf_dir())
    except:
        sys.exit(2)