Re: [Help-bash] how to find only the third command is valid?

[Greg Wooledge]

On Fri, Feb 05, 2016 at 06:01:24PM +0800, wk wrote:
> address@hidden test]# ./cmd -c "money sho"w
> address@hidden test]# ./cmd -c "money "show
> address@hidden test]# ./cmd -c "money show"
> 
>  Question:
>  I want to only allow case 3 :
>  ./cmd -n admin -p admin -c "money show"

There is literally no difference among these commands.  They all pass
*exactly* the same arguments, as you have already demonstrated with your
C program.

By the way, you can do the same thing in sh:

#!/bin/sh
printf "%d args:" "$#"
printf " <%s>" "$@"
echo

imadev:~$ args "quotes" """"""are''''$'' fun
3 args: <quotes> <are> <fun>

When the shell parses a command, one of the steps it performs is called
"quote removal".  In my example above, the first word the parser sees
is <args>, so that becomes the name of the command.  The second word it
sees is <"quotes">, and after quote removal and other expansions (none
of which apply here), the second word becomes <quotes>, and that will
be the first argument of the command.  The third word the parser sees is
<""""""are''''$''> which has three different kinds of quoting.  After
removing all of those quotes, the word becomes <are> and that is the
second argument of the command.  The final word is <fun> and needs no
further processing.

There is no way to differentiate among

./cmd -c "money sho"w
./cmd -c "money "show
./cmd -c "money show"

As far as the shell is concerned, they are all identical.  Every one of
them parses the final word as <money show>.

What exactly are you trying to achieve, and WHY?