python program to return exit code 0 if passes and 1 if fails

Question

I have a text file that includes a word like "test". Now what i am trying to do is using python i am opening that text file and search for that word. If the word exists then the python program should return exit code 0 or else 1.

This is the code that i have written that returns 0 or 1.

word = "test"

def check():
    with open("tex.txt", "r") as file:
        for line_number, line in enumerate(file, start=1):  
            if word in line:

                return 0
            else:
                return 1

print(check())

output

0

what I want is I want to store this exit code in some variable so that i can pass this to a yaml file. Can someone guide me here how can i store this exit code in a variable? thanks in advance

Asked By: Zykxpython

||

Source

Answer 1

I think you are looking for sys.exit(), but since you edited your question, I am not sure anymore. Try this:

import sys

word = "test"

def check():
    with open("tex.txt", "r") as file:
        for line_number, line in enumerate(file, start=1):  
            if word in line:
                return 0
     
    return 1

is_word_found = check() # store the return value of check() in variable `is_word_found`
print(is_word_found) # prints value of `is_word_found` to standard output
sys.exit(is_word_found) # exit the program with status code `is_word_found`

BTW: as @gimix and @VPfB mentioned, your check() function did return a value immediately after it checked the first line of your file. I included a fix for that, now 1 is returned only if the word is not found is any line of your file.

Answered By: Guillaume

python program to return exit code 0 if passes and 1 if fails

Question:

Answers: