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

Answer: 
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.
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.