Find files older than X days and output them by their size

files

I am trying to get the files older than a number of days and list them in descending order based on their size with all their information (size, full path etc – something similar that is provided by ls).

While I am able to locate files older with:

find . -mtime +10

I am not able to list of the desired information.

Best Answer

Provided that your file paths do not contain newline characters:

find . -mtime +10 -printf "%s %n %m %u %g %t %p" \( \
  -type l -printf ' -> %l\n' -o -printf '\n' \) | sort -k1,1 -n

See find manual, section Actions.

%s File's size in bytes.
%n Number of hard links to file.
%m File's permission bits (in octal).
%u File's user name, or numeric user ID if the user has no name.
%g File's group name, or numeric group ID if the group has no name.
%t File's last modification time in the format returned by the C ctime function.
%p File's name.
%l Object of symbolic link (empty string if file is not a symbolic link).

BTW: Note that POSIX find manual doesn't specify most of the above actions.

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If you want to delete all the files, then, on a GNU system, you could do:

cd -P -- "$destdir" &&
  printf '%s\0' * | # print the list of files as zero terminated records
    sort -Rz |      # random sort (shuffle) the zero terminated records
    xargs -r0 rm -f # pass the input if non-empty (-r) understood as 0-terminated
                    # records (-0) as arguments to rm -f

If you want to only delete a certain number of those matching a regexp you'd insert something like this between the sort and xargs:

awk -v RS='\0' -v ORS='\0' -v n=1024 '/regexp/ {print; if (--n == 0) exit}'

With zsh, you could do:

shuffle() REPLY=$RANDOM
rm -f file_<->_[a-d].bin(.+shuffle[1,1024])

Best Answer

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