Re: [Help-bash] Root access in subshell

[Bob Proulx]

Richard Taubo wrote:
> Seth David Schoen wrote:
> > Richard Taubo writes:
> > 
> >> I was actually trying to do the following (using $ as the prompt):
> >> $ OldIFS=$IFS; IFS=$'\n'; \
> >> for i in $(cat /etc/passwd); do \
> >> split_word=":"; my_user="${i%%$split_word*}";\
> >> if [[ "$my_user" != "root" && "$my_user" != “bin" ]]; then \
> >> printf "%s\n" "------- Finding files by user: $my_user";\
> >> find / -user $my_user -not -path "/proc/*" -print -quit;\
> >> fi;\
> >> done;\
> >> IFS=$OldIFS;

You say "Finding files" but doesn't the -print -quit only print one
file and then quit?

> > If you're up for using external Unix commands instead of shell
> > builtins, you can make this a bit more concise:
> > 
> > egrep -v '^(root|bin):' /etc/passwd | cut -d: -f1 | while read i; do
> > printf "%s\n" "------- Finding files by user: $i"
> > find / -user "$i" -not -path "/proc/*" -print -quit
> > fi
> > 
> > You could also save one byte (and one subprocess) by using awk:
> > 
> > awk -F: '!/^(root|bin):/ {print $1}' /etc/passwd
> 
> Thanks, those are good alternatives!

I would prefer Seth's awk solution.  With -F: it just works so nicely.

But reading the /etc/passwd file can be done internally to the shell
too.  Here is an example.

  #!/bin/sh
  IFS=:
  while read user pass uid gid gecos home shell; do
    printf "%s\n" "------- Finding file by user: $user"
    find / -user "$user" -not -path "/proc/*" -print -quit
  done < /etc/passwd
  exit 0

I might be inclined to use find -xdev though.  It all depends.
Definitely in an nfs environment I would.

Bob