python get command line arguments
import sys
print("This is the name of the script:", sys.argv[0])
print("Number of arguments:", len(sys.argv))
print("The arguments are:" , str(sys.argv))
#Example output
#This is the name of the script: sysargv.py
#Number of arguments in: 3
#The arguments are: ['sysargv.py', 'arg1', 'arg2']python sys.argv
def add(a, b):
return a + b
"""
You need to execute the script add.py as follows:
'python add.py 5 2'
The 'sys.argv[0]' is the name of your script,
so you need to get the second and the third one.
"""
print(add(int(sys.argv[1]), int(sys.argv[2])))
"""
With the command: 'python add.py 5 2',
this python script will returns 7
"""python sys.argv exception
try:
filename = sys.argv[1]
except IndexError:
print "You did not specify a file"
sys.exit(1)python sys
import sys
print('Python version you are using')
print(sys.version)
print('Informantion')
print(sys.version_info)how to take a list as input in python using sys.srgv
argv = sys.argv[1:]Copyright © 2021 Codeinu
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