sys.argv[0] always returns nothing

Question

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

There is no way to fix it, because there is nothing to fix? What were you hoping to make it return? — Paul Rooney, Jan 04 '19 at 02:41

score 2 · Accepted Answer · answered Jan 04 '19 at 02:35

2

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.

answered Jan 04 '19 at 02:35

U13-Forward

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You're right. Thank you! – Xiaoyu Jan 04 '19 at 03:18

score 0 · Answer 2 · edited Jan 04 '19 at 06:50

0

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.

edited Jan 04 '19 at 06:50

Prashant Pokhriyal

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answered Jan 04 '19 at 02:41

franckisses

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Test that and see what you get. – U13-Forward Jan 04 '19 at 02:43
Thank you. I did run the code in a shell. But sys.argv[0] still can't get the file name for me. Both sys.argv[1] and sys.argv[2] return: IndexError: list index out of range. So I think sys.argv is still empty. – Xiaoyu Jan 04 '19 at 02:53
@Xiaoyu You should see mine. – U13-Forward Jan 04 '19 at 02:55
otherwise.you should take a screenshot,let me see. – franckisses Jan 04 '19 at 09:16

sys.argv[0] always returns nothing

2 Answers2