用su模拟sudo的行为

Question

I'm trying to write a wrapper around su to make it more like sudo , so that su_wrapper foo bar baz == su -c "foo bar baz" .

However, I'm not sure how to approach this problem. I came up with this:

su_wrapper ()
{
  su -c "$@"
}

However, in the above, only a single argument can be there; this fails with multiple arguments (as su sees them as its own arguments).

There's also another problem: since the argument is passed through the shell, I think I must explicitly specify the shell to avoid other problems. Perhaps what I want to do could be expressed in pseudo-bash(!) as su -c 'bash -c "$@"' .

So, how could I make it accept multiple arguments?

Answer 1

Use printf "%q" to escape the parameters so they can be used as string input to your function:

su_wrapper() {
    su -s /bin/bash -c "$(printf "%q " "$@")"
}

Unlike $* , this works even when the parameters contain special characters and whitespace.

Answer 2

You need $* and not $@ :

su_wrapper() {
    local IFS=' '
    su -c "$*"
}

See the Special Parameters section in the Bash Reference Manual for the difference between $* and $@ .

I added local IFS=' ' just in case IFS is set to something else (after reading what $* does, it should be clear why you want to make sure that IFS is set to a space).