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']pass argument to a py file
import sys
def hello(a,b):
print "hello and that's your sum:", a + b
if __name__ == "__main__":
a = int(sys.argv[1])
b = int(sys.argv[2])
hello(a, b)
# If you type : py main.py 1 5
# It should give you "hello and that's your sum:6"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)how to take a list as input in python using sys.srgv
argv = sys.argv[1:]argv in python
#argv in python
import sys
print 'Number of arguments:', len(sys.argv), 'arguments.'
print 'Argument List:', str(sys.argv)
#command in cmd :
$ python test.py arg1 arg2 arg3
# producing following result :
Number of arguments: 4 arguments.
Argument List: ['test.py', 'arg1', 'arg2', 'arg3']Copyright © 2021 Codeinu
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