Cfrgtkky

for x in $(ls -ll <path to files> | awk '{ print $3,$4 }' | tail -n +2) ; do 

  if [ "${x}" != "root" ] ; then

    echo "Fail"

    break

  else

    echo "Pass"

 fi

done

Now, this prints "Pass" for every file it finds. I want to print "Pass" if all files are owned by root, and print "Fail" if any user or group in list is not root.

If you want to find out if all files in your are owned by root and belong to group root, use find:

find <path to files> ! -user root -or ! -group root -print

If anything is returned, that file either is not owned by root, or does not belong to the group root. You can then put that into a conditional clause to print out Pass or Fail.

[[ "$(find <path to files> ! -user root -or ! -group root -print)" == "" ]] && echo "Pass" || echo "Fail"

Hope this helps.

First, you shouldn't parse the output of ls and its variations. You can go about this using stat:

$ stat -c%U-%G ./*

tomasz-tomasz

tomasz-tomasz

tomasz-tomasz

As you can see, the result is a reliable list of two words concatenated, which you can operate on to get the result wanted. Put it into a loop, and there you go:

PASS=true

for i in $(stat -c%U-%G ./*); do

    if ! [[ "$i" == root-root ]]; then

        PASS=false; break

    fi

done

if "$PASS"; then

    echo Pass

else

    echo Fail

fi

The value of i needs to be root-root for the loop to get to its end with the switch unchanged.

Replace ./* with the_dir/* to test another location.

The - separator is needed, because, as Grump noted in the comments, The string comparison may fail if the file is owned by 'roo' and in the group 'troot', so a separator would still be a good thing.

Familiarise yourself with this: Why *not* parse `ls` (and what do to instead)?

How about

[ 1 = $({ echo root:root; stat -c"%U:%G" *; } | sort -u | wc -l) ] && echo PASS || echo FAIL

EDIT: or

[ -z $(stat -c"%U:%G" * | grep -vm1 root:root) ] && echo PASS || echo FAIL

The way I would do it is probably

found=$(find <path to files> -maxdepth 1 -not ( -user root -group root ) -printf "x")

found=${found:+Fail}

echo ${found:=Pass}

However, the simplest way of altering your script is:

found="Pass"



for x in $(ls -llA <path to files> | awk 'FNR>2{ print $3,$4 }' )

do 

   if [ "${x}" != "root" ]

   then

       found="Fail"

       break

   fi

done



echo $found

here I've added the A flag to catch files that begin with a "."

Doing it this way, however, is not a good idea as filenames containing a newline will cause unexpected results.

The output of ls is risky in command lines. I suggest using find for this purpose in the shellscript.

• The parameters to find -printf are described in man find.


• The standard output is piped to grep and the exit status is stored in norootfile.


• The output is also written to a temporary file, and the number of lines is counted and stored in numfile (the number of files found).


• You can use the option -v 'verbose' to get more details in the output from the shellscript.

If you want to search also hidden files use find . instead of find *

If you don't want to search in subdirectories, use -maxdepth 1 in the find command line.

#!/bin/bash



if [ "$1" == "-h" ]

then

 echo "Usage: $0 -h  # this help text"

 echo "       $0 -v  # verbose output"

 exit

fi



tmpfil=$(mktemp)



find * -xtype f -printf "%u:%g  %pn" | tee "$tmpfil" | grep -v '^root:root' > /dev/null



norootsfile=$?

numfile=$(wc -l "$tmpfil")

#cat "$tmpfil"



if [ ${numfile%% *} -eq 0 ]

then

 echo "No file found; check the current directory"

elif [ $norootsfile -eq 0 ]

then

 echo "Fail"

 if [ "$1" == "-v" ]

 then

  echo "----- Found some file(s) not owned or grouped by root"

  echo "user:group file-name --------------------------------"

  grep -v '^root:root' "$tmpfil"

 fi

else

 echo "Pass"

 if [ "$1" == "-v" ]

 then

  echo "----- Found only files owned or grouped by root"

 fi

fi

rm "$tmpfil"