[python] sys.argv[1] meaning in script

I'm currently teaching myself Python and was just wondering (In reference to my example below) in simplified terms what the sys.argv[1] represents. Is it simply asking for an input?

#!/usr/bin/python3.1

# import modules used here -- sys is a very standard one
import sys

# Gather our code in a main() function
def main():
  print ('Hello there', sys.argv[1])
  # Command line args are in sys.argv[1], sys.argv[2] ..
  # sys.argv[0] is the script name itself and can be ignored

# Standard boilerplate to call the main() function to begin
# the program.
if __name__ == '__main__':
  main()

sys.argv is a attribute of the sys module. It says the arguments passed into the file in the command line. sys.argv[0] catches the directory where the file is located. sys.argv[1] returns the first argument passed in the command line. Think like we have a example.py file.

example.py

import sys # Importing the main sys module to catch the arguments
print(sys.argv[1]) # Printing the first argument

Now here in the command prompt when we do this:

python example.py

It will throw a index error at line 2. Cause there is no argument passed yet. You can see the length of the arguments passed by user using if len(sys.argv) >= 1: # Code. If we run the example.py with passing a argument

python example.py args

It prints:

args

Because it was the first arguement! Let's say we have made it a executable file using PyInstaller. We would do this:

example argumentpassed

It prints:

argumentpassed

It's really helpful when you are making a command in the terminal. First check the length of the arguments. If no arguments passed, do the help text.

Adding a few more points to Jason's Answer :

For taking all user provided arguments : user_args = sys.argv[1:]

Consider the sys.argv as a list of strings as (mentioned by Jason). So all the list manipulations will apply here. This is called "List Slicing". For more info visit here.

The syntax is like this : list[start:end:step]. If you omit start, it will default to 0, and if you omit end, it will default to length of list.

Suppose you only want to take all the arguments after 3rd argument, then :

user_args = sys.argv[3:]

Suppose you only want the first two arguments, then :

user_args = sys.argv[0:2]  or  user_args = sys.argv[:2]

Suppose you want arguments 2 to 4 :

user_args = sys.argv[2:4]

Suppose you want the last argument (last argument is always -1, so what is happening here is we start the count from back. So start is last, no end, no step) :

user_args = sys.argv[-1]

Suppose you want the second last argument :

user_args = sys.argv[-2]

Suppose you want the last two arguments :

user_args = sys.argv[-2:]

Suppose you want the last two arguments. Here, start is -2, that is second last item and then to the end (denoted by ":") :

user_args = sys.argv[-2:]

Suppose you want the everything except last two arguments. Here, start is 0 (by default), and end is second last item :

user_args = sys.argv[:-2]

Suppose you want the arguments in reverse order :

user_args = sys.argv[::-1]

Hope this helps.

sys .argv will display the command line args passed when running a script or you can say sys.argv will store the command line arguments passed in python while running from terminal.

Just try this:

import sys
print sys.argv

argv stores all the arguments passed in a python list. The above will print all arguments passed will running the script.

Now try this running your filename.py like this:

python filename.py example example1

this will print 3 arguments in a list.

sys.argv[0] #is the first argument passed, which is basically the filename. 

Similarly, argv1 is the first argument passed, in this case 'example'

A similar question has been asked already here btw. Hope this helps!

Just adding to Frederic's answer, for example if you call your script as follows:

./myscript.py foo bar

sys.argv[0] would be "./myscript.py" sys.argv[1] would be "foo" and sys.argv[2] would be "bar" ... and so forth.

In your example code, if you call the script as follows ./myscript.py foo , the script's output will be "Hello there foo".

sys.argv[1] contains the first command line argument passed to your script.

For example, if your script is named hello.py and you issue:

$ python3.1 hello.py foo

or:

$ chmod +x hello.py  # make script executable
$ ./hello.py foo

Your script will print:

Hello there foo

sys.argv is a list containing the script path and command line arguments; i.e. sys.argv[0] is the path of the script you're running and all following members are arguments.

sys.argv is a list.

This list is created by your command line, it's a list of your command line arguments.

For example:

in your command line you input something like this,

python3.2 file.py something

sys.argv will become a list ['file.py', 'something']

In this case sys.argv[1] = 'something'

To pass arguments to your python script while running a script via command line

python create_thumbnail.py test1.jpg test2.jpg

here, script name - create_thumbnail.py, argument 1 - test1.jpg, argument 2 - test2.jpg

With in the create_thumbnail.py script i use

sys.argv[1:]

which give me the list of arguments i passed in command line as ['test1.jpg', 'test2.jpg']