Passing empty arguments to sudo -i

问题

Here is an example bash script that I would like to run using sudo -i:

#!/bin/bash

echo "arg 1: $1"
echo "arg 2: $2"

When I run this command normally with one empty argument, it runs as expected:

$ /tmp/args.sh "" two
arg 1:
arg 2: two

With plain sudo, I get the expected result:

$ sudo /tmp/args.sh "" two
arg 1:
arg 2: two

However if I use -i (to pick up the user's shell and login scripts), suddenly the first argument disappears:

$ sudo -i /tmp/args.sh "" two
arg 1: two
arg 2:

And I cannot figure out any way to quote or escape the first, empty argument.

Note: I have figured out a workaround by writing /tmp/args.sh "" two to a file and then executing that with sudo -i, but I was wondering if there is any way to achieve this directly from the command line.

回答1:

This works for me:

sudo -i bash -c '/tmp/args.sh "" two'

回答2:

Quote your empty double-quoted arguments with single quotes, like this:

$ sudo -i /tmp/args.sh '""' two

来源：https://stackoverflow.com/questions/27892812/passing-empty-arguments-to-sudo-i