I have been working on a script mixture of bash and python script. The bash script can receive unknown count input arguments. For example :
tinify.sh test1.jpg test2.jpg test3.jpg .....
After the bash receives all, it pass these arguments to tinify.py. Now I have come out two ways to do that.
Loop in bash and call python tinify.py testx.jpg
In another word, python tinify test1.jpg then python tinify test2.jpg, finaly python tinify test3.jpg
Pass all arguments to tinify.py then loop in python
But there is a problem, I want to filter same parameter for example if user input tinify.sh test1.jpg test1.jpg test1.jpg , I only want tinify.sh test1.jpg.So I think it's easier to do in second way because python may be convenient.
How can I do to pass all arguments to python script? Thanks in advance!
In addition to Chepner's answer above:
#!/bin/bash
tinify.py "$#"
within python script, tinify.py:
from sys import argv
inputArgs = sys.argv[1:]
def remove_duplicates(l):
return list(set(l))
arguments=remove_duplicates(inputArgs)
The list arguments will contain the arguments passed to python script (duplicated removed as set can't contain duplicated values in python).
You use $# in tinify.sh
#!/bin/bash
tinify.py "$#"
It will be far easier to eliminate duplicates inside the Python script that to filter them out from the shell. (Of course, this raises the question whether you need a shell script at all.)
A python program can accept any number of command line arguments, using sys.argv — just remember that sys.argv[0] is the name of the script
and actual arguments are contained in sys.argv[1:]
$ cat test_args.py
from sys import argv
prog_name = argv[0]
print('Program name:', prog_name)
for arg in argv[1:]:
print(arg)
$ python test_args.py a b 'c d'
Program name: test_args.py
a
b
c d
$
note that an argument containing spaces must be quoted according to the shell syntax.
Your file tinify.py should start with the following (if you have two arguments):
import sys
arg1, arg2 = sys.argv[1], sys.argv[2]
sys.argv[0] is the name of the script itself. You can of course loop over sys.argv. Personally, i like to pass all the arguments as json objects, so afterwards I do json.loads()
Related
I know that I can run a python script from my bash script using the following:
python python_script.py
But what about if I wanted to pass a variable / argument to my python script from my bash script. How can I do that?
Basically bash will work out a filename and then python will upload it, but I need to send the filename from bash to python when I call it.
To execute a python script in a bash script you need to call the same command that you would within a terminal. For instance
> python python_script.py var1 var2
To access these variables within python you will need
import sys
print(sys.argv[0]) # prints python_script.py
print(sys.argv[1]) # prints var1
print(sys.argv[2]) # prints var2
Beside sys.argv, also take a look at the argparse module, which helps define options and arguments for scripts.
The argparse module makes it easy to write user-friendly command-line interfaces.
Use
python python_script.py filename
and in your Python script
import sys
print sys.argv[1]
Embedded option:
Wrap python code in a bash function.
#!/bin/bash
function current_datetime {
python - <<END
import datetime
print datetime.datetime.now()
END
}
# Call it
current_datetime
# Call it and capture the output
DT=$(current_datetime)
echo Current date and time: $DT
Use environment variables, to pass data into to your embedded python script.
#!/bin/bash
function line {
PYTHON_ARG="$1" python - <<END
import os
line_len = int(os.environ['PYTHON_ARG'])
print '-' * line_len
END
}
# Do it one way
line 80
# Do it another way
echo $(line 80)
http://bhfsteve.blogspot.se/2014/07/embedding-python-in-bash-scripts.html
use in the script:
echo $(python python_script.py arg1 arg2) > /dev/null
or
python python_script.py "string arg" > /dev/null
The script will be executed without output.
I have a bash script that calls a small python routine to display a message window. As I need to use killall to stop the python script I can't use the above method as it would then mean running killall python which could take out other python programmes so I use
pythonprog.py "$argument" & # The & returns control straight to the bash script so must be outside the backticks. The preview of this message is showing it without "`" either side of the command for some reason.
As long as the python script will run from the cli by name rather than python pythonprog.py this works within the script. If you need more than one argument just use a space between each one within the quotes.
and take a look at the getopt module.
It works quite good for me!
Print all args without the filename:
for i in range(1, len(sys.argv)):
print(sys.argv[i])
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()
You may have been directed here because you were asking about an IndexError in your code that uses sys.argv. The problem is not in your code; the problem is that you need to run the program in a way that makes sys.argv contain the right values. Please read the answers to understand how sys.argv works.
If you have read and understood the answers, and are still having problems on Windows, check if Python Script does not take sys.argv in Windows fixes the issue. If you are trying to run the program from inside an IDE, you may need IDE-specific help - please search, but first check if you can run the program successfully from the command line.
I would like to note that previous answers made many assumptions about the user's knowledge. This answer attempts to answer the question at a more tutorial level.
For every invocation of Python, sys.argv is automatically a list of strings representing the arguments (as separated by spaces) on the command-line. The name comes from the C programming convention in which argv and argc represent the command line arguments.
You'll want to learn more about lists and strings as you're familiarizing yourself with Python, but in the meantime, here are a few things to know.
You can simply create a script that prints the arguments as they're represented. It also prints the number of arguments, using the len function on the list.
from __future__ import print_function
import sys
print(sys.argv, len(sys.argv))
The script requires Python 2.6 or later. If you call this script print_args.py, you can invoke it with different arguments to see what happens.
> python print_args.py
['print_args.py'] 1
> python print_args.py foo and bar
['print_args.py', 'foo', 'and', 'bar'] 4
> python print_args.py "foo and bar"
['print_args.py', 'foo and bar'] 2
> python print_args.py "foo and bar" and baz
['print_args.py', 'foo and bar', 'and', 'baz'] 4
As you can see, the command-line arguments include the script name but not the interpreter name. In this sense, Python treats the script as the executable. If you need to know the name of the executable (python in this case), you can use sys.executable.
You can see from the examples that it is possible to receive arguments that do contain spaces if the user invoked the script with arguments encapsulated in quotes, so what you get is the list of arguments as supplied by the user.
Now in your Python code, you can use this list of strings as input to your program. Since lists are indexed by zero-based integers, you can get the individual items using the list[0] syntax. For example, to get the script name:
script_name = sys.argv[0] # this will always work.
Although interesting, you rarely need to know your script name. To get the first argument after the script for a filename, you could do the following:
filename = sys.argv[1]
This is a very common usage, but note that it will fail with an IndexError if no argument was supplied.
Also, Python lets you reference a slice of a list, so to get another list of just the user-supplied arguments (but without the script name), you can do
user_args = sys.argv[1:] # get everything after the script name
Additionally, Python allows you to assign a sequence of items (including lists) to variable names. So if you expect the user to always supply two arguments, you can assign those arguments (as strings) to two variables:
user_args = sys.argv[1:]
fun, games = user_args # len(user_args) had better be 2
So, to answer your specific question, sys.argv[1] represents the first command-line argument (as a string) supplied to the script in question. It will not prompt for input, but it will fail with an IndexError if no arguments are supplied on the command-line following the script name.
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.
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'
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".
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]
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.
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']
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.
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, argv[1] is the first argument passed, in this case 'example'.
Is it possible to somehow convert string which I pass via command line e.g
python myfile.py "s = list(range(1000))"
into module? And then...
#myfile.py
def printlist(s):
print(s)
?
Something like a timeit module, but with my own code.
python -m timeit -s "s = list(range(1000))" "sorted(s)"
Save the code
import sys
print(sys.argv)
as filename.py, then run python filename.py Hi there! and see what happens ;)
I'm not sure if I understand correctly, but the sys module has the attribute argv that is a list of all the arguments passed in the command line.
test.py:
#!/usr/bin/env python
import sys
for argument in sys.argv:
print(argument)
This example will print every argument passed to the script, take in mind that the first element sys.argv[0] will always be the name of the script.
$ ./test.py hello there
./test.py
hello
there
There are two options to send short code snippets similar to what you can do via the timeit module or certain other -m module option functionality and these are:
You can use either the -c option of Python e.g.:
python -c "<code here>"
Or you could use piping such as:
echo <code here> | python
You can combine multiple statements using the ; semicolon statement separator. However if you use a : such as with while or for or def or if then you cannot use the ; so there may be limited options. Possibly some clever ways around this limitation but I have yet to see them.
So to be more precise, what I am trying to do is :
read a full shell command as argument of my python script like : python myPythonScript.py ls -Fl
Call that command within my python script when I'd like to (Make some loops on some folders and apply the command etc ...)
I tried this :
import subprocess
from optparse import OptionParser
from subprocess import call
def execCommand(cmd):
call(cmd)
if __name__ == '__main__':
parser = OptionParser()
(options,args) = parser.parse_args()
print args
execCommand(args)
The result is that now I can do python myPythonScript.py ls , but I don't know how to add options. I know I can use parser.add_option , but don't know how to make it work for all options as I don't want to make only specific options available, but all possible options depending on the command I am running.
Can I use something like parser.add_option('-*') ? How can I parse the options then and call the command with its options ?
EDIT
I need my program to parse all type of commands passed as argument : python myScript.py ls -Fl , python myScript.py git pull, python myScript rm -rf * etc ...
OptionParser is useful when your own program wants to process the arguments: it helps you turn string arguments into booleans or integers or list items or whatever. In your case, you just want to pass the arguments on to the program you're invoking, so don't bother with OptionParser. Just pass the arguments as given in sys.argv.
subprocess.call(sys.argv[1:])
Depending on how much your program depends on command line arguments, you can go with simple route.
Simple way of reading command line arguments
Use sys to obtain all the arguments to python command line.
import sys
print sys.argv[1:]
Then you can use subprocess to execute it.
from subprocess import call
# e.g. call(["ls", "-l"])
call(sys.argv[1:])
This sample below works fine for me.
import sys
from subprocess import call
print(sys.argv[1:])
call(sys.argv[1:])
Is anyone able to tell me how to write a conditional for an argument on a python script? I want it to print "Argument2 Entered" if it is run with a second command line arguments such as:
python script.py argument1 argument2
And print "No second argument" if it is run without command line arguments, like this:
python script.py argument1
Is this possible?
import sys
if len(sys.argv)==2: # first entry in sys.argv is script itself...
print "No second argument"
elif len(sys.argv)==3:
print "Second argument"
There are many answers to this, depending on what exactly you want to do and how much flexibility you are likely to need.
The simplest solution is to examine the variable sys.argv, which is a list containing all of the command-line arguments. (It also contains the name of the script as the first element.) To do this, simply look at len(sys.argv) and change behaviour based on its value.
However, this is often not flexible enough for what people expect command-line programs to do. For example, if you want a flag (-i, --no-defaults, ...) then it's not obvious how to write one with just sys.argv. Likewise for arguments (--dest-dir="downloads"). There are therefore many modules people have written to simplify this sort of argument parsing.
The built-in solution is argparse, which is powerful and pretty easy-to-use but not particularly concise.
A clever solution is plac, which inspects the signature of the main function to try to deduce what the command-line arguments should be.
There are many ways to do this simple thing in Python. If you are interested to know more than I recommend to read this article. BTW I am giving you one solution below:
import click
'''
Prerequisite: # python -m pip install click
run: python main.py ttt yyy
'''
#click.command(context_settings=dict(ignore_unknown_options=True))
#click.argument("argument1")
#click.argument("argument2")
def main(argument1, argument2):
print(f"argument1={argument1} and argument2={argument2}")
if __name__ == '__main__':
main()
Following block should be self explanatory
$ ./first.py second third 4th 5th
5
$ cat first.py
#!/usr/bin/env python
import sys
print (len(sys.argv))
This is related to many other posts depending upon where you are going with this, so I'll put four here:
What's the best way to grab/parse command line arguments passed to a Python script?
Implementing a "[command] [action] [parameter]" style command-line interfaces?
How can I process command line arguments in Python?
How do I format positional argument help using Python's optparse?
But the direct answer to your question from the Python docs:
sys.argv -
The list of command line arguments passed to a Python script. argv[0] is the script name (it is operating system dependent whether this is a full pathname or not). If the command was executed using the -c command line option to the interpreter, argv[0] is set to the string '-c'. If no script name was passed to the Python interpreter, argv[0] is the empty string.
To loop over the standard input, or the list of files given on the command line, see the fileinput module.