Develop Reference

Passing a shell variable to inline python in a shell script

I have a simple shell script where i want to be able to pass variables to some inline python i will write. For example like this
funny=879
echo $funny
python -c "
print(f"hello {$funny}")
"
However this prints
879
File "<string>", line 2
print(fhello
^
SyntaxError: unexpected EOF while parsing
(pipeline) $
Any thoughts on what i could be doing wrong? I know i am setting the variable correct because when i do echo it prints out the variable so it is definitely set correct but for some reason python script is not able to use it.

It's because you're using outer double quotes.
python -c "print(f"hello {$funny}")"
Gets turned into:
python -c print(fhello {879})
So python is passed 2 separate strings.
The inner double quotes would need to be escaped in order to get passed through to python.
$ funny=879; python3 -c "print(f\"hello $funny\")"
hello 879
Instead of messing around with quoting - if you export your variables you can access them from python using the os.environ dict.
$ export funny=879; python -c 'import os; print(os.environ["funny"])'
879
You can use the var=value command syntax and omit the export (note the lack of a semicolon)
$ funny=879 fonny=978 python3 -c 'import os; print(os.environ["funny"], os.environ["fonny"])'
879 978

Pass parameter from command line in python

I have a simple task and want to run command from the line.
E.g.
python3 -c 'print(2*2)'
The issue is when I want to invoke a function and pass parameter to it. E.g I want to lower the string 'ABC'.
I use
python3 -c 'print(x="ABC",x.lower())'
Hence my question: how could I pass string value to function when invoke python from command line?

Use sys.argv to obtain command-line arguments:
$ python3 -c "import sys; print(sys.argv[1].lower())" HELLO
hello
$

Python fabric cant be executed from python using subproccess

In python I am trying to execute a fabfile and ,I get the below error from the subproccess output. I installed fabric using easy install. If I run the code from the command line it works. From python no go. I assume there is an issue with how I am using the Popen command?
/bin/sh: 1: fab: not found
Below is how I start fabric from python:
cmd = """fab -H 111.111.111.111 aws_bootstrap initial_chef_run:aws_server.json,aws-test,development -w """
os.chdir(fab_path) #change to the dir where the fabfile is located
p = subprocess.Popen(cmd, shell=True,stderr=subprocess.STDOUT,stdout=subprocess.PIPE)
PS I added the below to the Popen but get the below error:
executable="/bin/bash"
/bin/bash: fab: command not found
From the command line I get the below which means the terminal can find fab.
fab
Fatal error: Couldn't find any fabfiles!
Remember that -f can be used to specify fabfile path, and use -h for help.
Aborting.

Try to use the whole path for fab. To get the path to fab on your system, you can use which fab.
However, I can't think of any reason why calling fab from python might be better than using the execute function of fab:
from fabric.tasks import execute
import my_fabfile
r = execute(my_fabfile.aws_bootstrap, hosts=["root#%s" % '111.111.111.111])
Return value r will contain a dict with the hosts as key(s).

Interpreting python program line by line from makefile

I need to interpret a python program line by line. I am using -c option to python and have makefile like this.
all:
python -c
"print 'aa'
print 'bb'"
When I run it with make I get
python -c "print 'aa'
/bin/sh: -c: line 0: unexpected EOF while looking for matching `"'
/bin/sh: -c: line 1: syntax error: unexpected end of file
make: *** [all] Error 2
when I take out the same python lines and run from bash, it works fine. What could be the problem?

If your Makefile truly is
all:
python -c
"print 'aa'
print 'bb'"
I would expect to see more errors. With that makefile, make will first attempt to run python -c, which should generate errors like:Argument expected for the -c option. It will then abort and not even try to run the shell command "print 'aa'. You need line continuations and semi-colons.
all:
python -c \
"print 'aa'; \
print 'bb'"
The semi-colon is necessary because make strips all the newlines and passes the string python -c "print 'aa'; print bb'" to the shell (whatever SHELL is set to).

Every line of a make rule is executed in a different shell instance. You need to escape the newlines (with \) or put it all on one line.
Also the makefile snippet as given should be giving you an error about unexpected arguments to -c. Your error indicates that your snippet is actually:
all:
python -c "print 'aa'
print 'bb'"
Not that that changes anything.

Have a look at this question. I think your problem is that your program is spanning multiple lines, but your makefile isn't interpreting it that way. Adding slashes should clear that up.
Multiline bash commands in makefile

Bash not recognising python command

I have this command as part of a bash script
$(python -c "import urllib, sys; print urllib.unquote(sys.argv[0])", "h%23g")
But when I run it, I get this:
-bash: -c: command not found
As though bash has missed reading the python, and is thinking -c is the name of the command. Exactly the same happens when using backticks.
How can I make bash recognise the python?

the Python command is returning the string "-c" from your $(...) structure, which bash then tries to execute.
for example
python -c "import urllib, sys; print urllib.unquote(sys.argv[0])"
prints "-c", so you are essentially asking bash to interpret $(-c), for which the error is valid.

I think you want something like the following:
$(python -c "import urllib, sys; print urllib.unquote(sys.argv[1])" "h%23g")
This will result in h#g, if this is all you have on a line then it will also attempt to run a command called h#g, so I'm assuming you are actually using this as a part of a larger command.
The issue with your version is that sys.argv[0] is the -c from the command, and urllib.unquote('-c') will just return '-c'.
From the documentation on sys.argv:
If the command was executed using the -c command line option to the interpreter, argv[0] is set to the string '-c'.
Combining that with info from the man page (emphasis mine):
-c command
Specify the command to execute (see next section). This terminates the option list (following options are passed as arguments to the command).
So, when you use -c, sys.argv[0] will be '-c', the argument provided to -c is the script so it will not be included in sys.argv, and any additional arguments are added to sys.argv starting at index 1.