I have a bash script and tried very hard but i couldn't solve it please help

Shell Programming and Scripting
1

please can you help me with this script ( very very important )

what I'm trying is to write program that accepts list of user as its argument
1- If a user or more are given as arguments, the script should reset files permissions as follows:
a. Directory ~/share to 750 (if it exists).
b. All regular files inside ~/share to 744.
c. All directories inside ~/share to 750.
d. All other regular files in ~ to 600 and all other directories in ~ to 700.
e. files of other types are left as they are.

We assume that there is only one level of files and directories inside ~ (except for ~/share), and there is only one level of files and directories inside ~/share.

2- If no arguments are give, the script should identify normal users in the system (UID >= 500) and for each user, reset files permissions as described above.

3- At the end of the execution, the script must produce a file named report.txt that contains a list of all files in the user's home directory including full path, owner, group and permissions for each file.

please help me if you can

2

Look very much homework...

What have you done so far?

3

I am very new to Linux and this is what some friend of mine give me to test what I learn
and i tried very much to solve it but till now I couldn't do anything

4

All right, what have you tried then?

5

Start with "man find" especially -type and -path, but not -exec.

Something like this is very durable and fast, one line for each situation:

find ... -type d | xargs -n 101 chmod 765
6 
[ -d ~/share ] && find ~/share -type f -print0 | while read -d $'\0' file
do
  # if $file exists and is a regular file
  [ -f "$file" ] && chmod 744 "$file"
done
7

Easy to match to spec and comment (case is nice that way, but use balanced parentheses so vi % is not disabled):
(Note: files in ~ like .profile, code, scripts no longer executable)

find ~ |while read p
do
 case "$p" in
 (~/share)
   chmod 750 "$p"
   ;;
 (~/share/*)
   if [ -f "$p" ];then
     chmod 744 "$p"
   else
     if [ -d "$p" ];then
       chmod 750 "$p"
     fi
   fi
   ;;
 (*)
    if [ -f "$p" ];then
     chmod 600 "$p"
   else
     if [ -d "$p" ];then
       chmod 700 "$p"
     fi
   fi
   ;;
 esac
done
1 Like
8

and what about the second one
2- If no arguments are give, the script should identify normal users in the system (UID >= 500) and for each user, reset files permissions as described above.

---------- Post updated at 01:11 PM ---------- Previous update was at 12:17 PM ----------

and what about the second one
2- If no arguments are give, the script should identify normal users in the system (UID >= 500) and for each user, reset files permissions as described above.

9

I would do that in a wrapper, and call the separately tested child script from it. It simplifies testing of both. If you want, later one can be a function within the other.

cut, sed or awk could find the /etc/passwd fields you need, and selectively invoke the script. Somewhere, ensure the home dir exists or is gracefully logged and ignored!

sed '
  s/^[^:]*:[^:]*:\([^:]*\):[^:]*:[^:]*:\([^:]*\):.*/\1 \2/
  /^[0-9]\{0,2\} /d
  /*[0-4][0-9][0-9] /d
  / \/$/d
  / $/d
  s/^/your_script /
 ' /etc/passwd >all_script

Narrative: Pull fields 3 and 6, colon : separated, from /etc/passwd, now separated by a space; if 3, the user #, is 0 to 2 digits, or if it is three digits starting 0-4, it is too low. If the nominal home dir is '/', we don't want to do this. If there is no home dir, we don't want to try. Stick your script on the front of every line.

10

L will try it now .

11

Enjoy, but be careful!