sys.argv[0] always returns nothing

user_10849416 0
Viewed 1209

I was trying to use sys.argv[0] to get the name of the script but it returned nothing. I guess that was because no script name was passed to the Python interpreter but I had no idea how to fix it.

my code:

import sys
print ("This is the name of the script: ", sys.argv[0])
sys.argv[0]

outputs:

>>> import sys
>>> print ("This is the name of the script: ", sys.argv[0])
This is the name of the script:
>>> sys.argv[0]
''

Thank you

read-eval-print-loop sys argv python
2 Answers

Well, that's well expected,

you're running your code on the interpreter, which is not any module nor file, so sys.argv knows that and gives you an empty string.

It's a good sign.

If you run it in an actual module or file, it will work perfectly, as expected.

user_87083640

You should run your code in a shell, then the arguments fellow the python filename, just like: python3 test.py first second. then in the code, you can find the args in the file.

print(sys.argv[1]) -------> first<br>    
print(sys.argv[2]) -------> second

I hope it will be helpful.

user_108655020
Related