[SOLVED] Any way to know beforehand if SUDO is (going to be) needed?

courteous · 12-12-2010, 01:43 AM

I'm using virtual file-system in /proc/ to print out current working directory (CWD):

Code:

ls /proc/$PID/cwd # part of an argument to printf (full printf below)

Now, some processes complain about SUDO privileges (e.g. init: ls: cannot access /proc/1/cwd: Permission denied).
Is it possible to know beforehand if action/command will require SUDO? For example, is there a "maximum PID number", above which no process needs SUDO? Is the following snippet (that checks if user ran the script as SUDO) good enough for behorand-knowledge?

Code:

if [ $UID -eq 0 ] ; then
# do something

BTW, in following printf snippet with 2 arguments,

Code:

printf "%15s  %s\n"  CWD:  $((ls /proc/$PID/cwd) | tr '\n' ' ')

if ls returns error cannot open directory /proc/1/fd: Permission denied it outputs this error before "CWD:", like so:

Code:

ls: cannot access /proc/1/cwd: Permission denied
            CWD:

Why this happens?

stress_junkie · 12-12-2010, 05:02 AM

You've asked two questions. I will address the first one.

Security settings in Linux are often tweaked by each distribution. Therefore when you are looking around the guts of the system you may have a command work on one distribution and fail on another due to security.

You can know in advance if some information in /proc is available by performing an "ls -l" to view the permission settings on the object that you want to access.

Regarding your second question I believe that you get the error on the second argument before the first argument is printed because you have specified a format string. I believe that the printf function in bash, which is different from the printf external command, tries to format the entire output before it prints the first result. So printf evaluates CWD and succeeds, then it evaluates the second argument and fails and gives an error, then it proceeds to print whatever was successfully processed.

You can find more information using the man pages.

Code:

man bash

Once you are in the man page for bash you can look for the printf command.

But, I could be wrong. I'm no cli expert.

courteous · 12-12-2010, 06:24 AM

Quote:

Originally Posted by stress_junkie

You can know in advance if some information in /proc is available by performing an "ls -l" to view the permission settings on the object that you want to access.

Indeed, I could use "ls -l" and then check if ( PID's folder owner != $USER ): if yes, then go ahead; if no, print error.

How do I use grep on "ls -l". For example, to find init #1 folder, this works:

Code:

ls | grep ^1$

EDIT: I get it, the last two CODEs don't return anything because grep operates on the whole "row" from "ls -l" (access modifiers, owner, etc.). Now, how can do you cut from "ls -l" only folder's name?

EDIT2: OMG, this works (prints folder names), but this surely isn't the "right way" to do it:

Code:

ls -l | cut -d' ' -f35

Why doesn't the default delimiter (TAB) work with "ls -l" output?!

unSpawn · 12-12-2010, 07:27 AM

An easy way to retrieve directory ownership would be 'stat -c %u /proc/$PID/cwd'. The easiest way to resolve a symbolic link would be to use 'readlink -f /proc/$PID/cwd'. Combining that you could
'[ `stat -c %u /proc/$PID/cwd` -eq $UID ] && { DISPLAY_CWD=$(readlink -f /proc/$PID/cwd); echo "CWD: ${DISPLAY_CWD:=noneofyourbusiness}"; };'.

rob.rice · 12-12-2010, 11:53 PM

MAN UP !!
you don't need no stinking sudo !
just log in as root