how to use sys.argv in python for exe file?

Asked Mar 02 '23 at 16:13

Active Mar 02 '23 at 16:38

Viewed 68 times

I want to use the following code so that when I run the exe file in CMD, I enter the name of my program, for example the number 2 or 3, and my code creates a file, otherwise it gives me the value "no" no no" but every time I run it, it gives me the following error

import sys

def test(numder):
    x =numder
    if x ==2:
        open("E:\myfile1.txt", "2")
        return "myfile1 crated"
    elif x ==3:
        open("E:\myfile2.txt", "3")
        return "myfile2 crated"
    return "no no no "

n= int(sys.argv[1])
print("n is: ",n)
x=test(n)
print(x)

asked Mar 02 '23 at 16:13

Navrang

1

Is `"2"` a valid `mode`? https://docs.python.org/3/library/functions.html#open – JonSG Mar 02 '23 at 16:19
1

@JonSG, fair, reopened. – Charles Duffy Mar 02 '23 at 16:38
2

@Navrang, what do you expect the `"2"` or `"3"` you pass to `open()` to _do_? – Charles Duffy Mar 02 '23 at 16:39

how to use sys.argv in python for exe file?

0 Answers0