scrip to change directory/file permissions

ramesh14 · 09-11-2008, 04:39 AM

I have a directory which contains some directories and files. and the sub-dirs contains some more dirs and files. All the directories and files inside the main directory are having different permissions. I want to give 0755 for all dirs and 0644 for all files in the main dir. Changing permissions for each and every thing is a critical job. This bash shell script if for a single dir or file (example: polo).

Code:

if [ -d polo ]
   chmod 0755 polo;
elif [ -f polo ]
   chmod 0644 polo;
else
   echo " polo is not a dir or file";

Further i am not understanding. I have to perform this action many time when ever i copy contents. Can any body please help me?

i92guboj · 09-11-2008, 05:02 AM

Quote:

Originally Posted by ramesh14

I have a directory which contains some directories and files. and the sub-dirs contains some more dirs and files. All the directories and files inside the main directory are having different permissions. I want to give 0755 for all dirs and 0644 for all files in the main dir. Changing permissions for each and every thing is a critical job. This bash shell script if for a single dir or file (example: polo).

Code:

if [ -d polo ]
   chmod 0755 polo;
elif [ -f polo ]
   chmod 0644 polo;
else
   echo " polo is not a dir or file";

Further i am not understanding. I have to perform this action many time when ever i copy contents. Can any body please help me?

This should work. Just substitute /path/to/ by the correct path, and all the files and dirs under that node will be processed.

Code:

find /path/to/ \( -type d -exec echo chmod 0755 '{}' \; -o -type f -exec chmod 0644 '{}' \; \)

If you want, you could as well do a bash function, for example, try putting this into your ~/.bashrc and ~/.bash_profile files:

Code:

# Function to fix permissions
function fixperms () {
  if [ -z "$1" ]
  then
    echo "Usage: fixperms base_directory [file permissions] [dir_permissions]"
    return
  fi
  if [ -z "$2" ]; then local FILE_PERMS=0644; else FILE_PERMS="$2"; fi
  if [ -z "$3" ]; then local DIR_PERMS=0755; else DIR_PERMS="$3"; fi
  echo "* Using $FILE_PERMS for files and $DIR_PERMS for directories."
  if [ -d "$1" ] || [ -f "$1" ]
  then
    find "$1" \( \
      -type d -exec chmod "$DIR_PERMS" '{}' \; -o \
      -type f -exec chmod "$FILE_PERMS" '{}' \; \
    \)
  else
    echo "! $1 not found."
  fi
}

This will be quite a time saver if you do this a lot. Now you can reload your shell (just close your terminal and open a new one, run exec bash or re-login if you wish) so the changes takes effect. From now on, you should be able to just run

Code:

$ fixperms /any/dir/you/want

Probably much easy to handle than the sentence I wrote above.

EDIT: I found an improved version of this function on my home and pasted it instead. It's basically the same with a couple of checks and basic help.

jschiwal · 09-11-2008, 05:37 AM

Working interactively, it may be easier to use the find command twice.

find /path/to/dir -type d -exec chmod 0755 '{}' \;
find /path/to/dir -type f -exec chmod 0644 '{}' \;

You do want to chmod the directories first in the case where you don't have read or executable rights for a directory. That will allow find to enter directories when you chmod the files.

vikas027 · 09-11-2008, 09:05 AM

Quote:

Originally Posted by jschiwal

find /path/to/dir -type d -exec chmod 0755 '{}' \;
find /path/to/dir -type f -exec chmod 0644 '{}' \;
.

This is the best solution.

jan61 · 09-11-2008, 01:35 PM

Moin,

Quote:

Originally Posted by vikas027

This is the best solution.

no, I don't think so ;-) You're starting a separate process for each file or directory. There's xargs to avoid this:

Code:

find /path/to/dir -type d -print | xargs chmod 0755
find /path/to/dir -type f -print | xargs chmod 0644

And if you have file names including blanks, line feeds and other special characters, use:

Code:

find /path/to/dir -type d -print0 | xargs -0 chmod 0755
find /path/to/dir -type f -print0 | xargs -0 chmod 0644

Jan