Bash无法识别python命令

Question

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 ?

Answer 1

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.

Answer 2

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.