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