CODEWITHSUNDEEP
pythonbash
ACarter
ACarter

Reputation: 5697

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?

Upvotes: 1

Views: 401

Answers (2)

Andrew Clark
Andrew Clark

Reputation: 208465

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.

Upvotes: 3

Matt W
Matt W

Reputation: 476

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.

Upvotes: 6

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