I'd like to check whether a program is installed on an UNIX system.
I could use commands like:
command -v
,hash
,type
,which
......and all of them are already mentioned in this answer.
However, none of them work if I want to test, as a normal user, whether I or any superuser can run given command.
Here's an example of what I mean:
dummy:~$ command -v poweroff; echo $?
1
dummy:~$ su
root:~# command -v poweroff; echo $?
/sbin/poweroff
0
As you see, normal user didn't find out about the existence of the poweroff
command. Note that dummy users can freely see what's in /sbin
anyway.
Use the command -v Command to Check if a Command Exists in Bash. The command -v is a built-in function in all POSIX systems and Bash. This function checks if a command exists as it returns the valid path for that command if it does exist and returns NULL if it does not.
You can use the posix compatible command to check, if the command exists from a bash script or not. if it returns >0 when the command is not found or an error occurs. similary you can use type and hash .
Now, every command run in bash shell returns a value that's stored in the bash variable “$?”. To get the value, run this command. $ echo $? If a command succeeded successfully, the return value will be 0.
bash [filename] runs the commands saved in a file. $@ refers to all of a shell script's command-line arguments. $1 , $2 , etc., refer to the first command-line argument, the second command-line argument, etc. Place variables in quotes if the values might have spaces in them.
The reason the commands you tried do not work is that they only look for executables in $PATH
variable. First, let's test our hypothesis.
dummy:~$ mkdir test
dummy:~$ cd test
dummy:~/test$ echo '#!/bin/sh' >test.sh
dummy:~/test$ chmod +x test.sh
dummy:~/test$ cd
dummy:~$ command -v test.sh
dummy:~$ PATH+=:/home/dummy/test/
dummy:~$ command -v test.sh
/home/dummy/test/test.sh
This confirms my statement above.
Now, let's have a look what $PATH
looks like for different users:
dummy:~$ echo $PATH
/usr/local/bin:/usr/bin:/bin:/usr/local/games:/usr/games
dummy:~$ su
root:~# echo $PATH
/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin
So in order to check whether given command is available to given user (in your question, namely: root), you need to know his $PATH
environment variable.
Values of such environment variables on Debian can be usually found in /etc/profile
and in /etc/environment/
files. There is no easy way to get these values by fishing them out of files.
The most basic solution is to temporarily add known directories to your $PATH
variable and then use command -v
:
dummy~$ OLDPATH=$PATH
dummy~$ PATH=$OLDPATH:/sbin:/usr/sbin/:/usr/local/sbin/
dummy~$ command -v poweroff
/sbin/poweroff
dummy~$ PATH=$OLDPATH
There is one problem with this solution: if you want to be portable, you don't really know what are the folders that you should concatenate. In most cases this approach should be sufficient, though.
What you can do instead is to write a script program that makes use of setuid bit. Setuid bit is a somewhat hidden feature of Linux operating systems that allows programs to be executed on their owner privileges. So you write a program that executes some commands like superuser would, except that it can be run by normal users. That way you can see output of command -v poweroff
like a root would do.
Unfortunately, stuff that uses shebang can't have setuid bit, so you cannot create a shell script for this and you need a program in C. Here's an example program that would do the job:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <unistd.h>
int main(int argc, char** argv)
{
if (argc <= 1)
{
fprintf(stderr, "No arguments.\n");
return 1;
}
//validate the argv
char* prog = argv[1];
int i;
for (i = 0; i < strlen(prog); i ++)
{
if (prog[i] < 'a' || prog[i] > 'z')
{
fprintf(stderr, "%s contains invalid characters (%c), exiting.", prog, prog[i]);
return 1;
}
}
//here's the `which` command. We start it in new interactive shell,
//since this program inherits environment variables from its
//parent shell. We need to start *new* shell that will initialize
//and overwrite existing PATH environment variable.
char* command = (char*) malloc(strlen(prog) + 30);
if (!command)
{
fprintf(stderr, "No memory!\n");
return 1;
}
sprintf(command, "bash -cli 'command -v %s'", prog);
int exists = 0;
//first we try to execute the command as a dummy user.
exists |= system(command) == 0;
if (!exists)
{
//then we try to execute the command as a root user.
setuid(0);
exists |= system(command) == 0;
}
return exists ? 0 : 1;
}
Security note: the version above has very simple argument validation (it lets through only strings matching ^[a-z]*$
). Real program should probably include better validation.
Suppose we saved the file in test.c
. We compile it and add setuid bit:
root:~# gcc ./test.c -o ./test
root:~# chown root:root ./test
root:~# chmod 4755 ./test
Note that chown
goes before chmod
. The 4
before usual 755
pattern is the setuid bit.
Now we can test the program as a normal user.
dummy:~$ ./test ls; echo $?
alias ls='ls -vhF1 --color=auto --group-directories-first'
0
dummy:~$ ./test blah; echo $?
1
dummy:~$ ./test poweroff; echo $?
/sbin/poweroff
0
And best of all - it's portable enough to work on cygwin with no problems. :)
The real answer is that you can't satisfy the "any superuser" aspect IF you mean: "Does this command appear in the search path of any user with sudo access?". The reason is that you'd have to run each user's startup scripts to find out what his search path ends up being - many users will include their own ~/bin or ~/pod/abi/x86_64-ubu-1204/bin or whatever and god knows what else (/afs//bin, anyone?) - and many startup scripts have side effects that could turn the whole thing into a real mess, including generating logs, starting up daemons of various kinds, and so on. You'll really be in trouble if one of those users' startup scripts tries to run your new command itself, since they'll then recurse and inflict a denial-of-service attack on your own system.
What you can test more safely is:
On a larger system with thousands of users, the "anyone" option isn't practical. Mature sites will NOT want a root-capable command running arbitrary users' scripts, either, even those of sudo-enabled, generally trustworthy admins.
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