How can I count the number of subdirectories that don't have the execute bit set?

This is my attempt but is there a better or more elegant way?

count=0; for d in */; do [[ -d $d && ! -x $d ]] && ((   count )); done
printf %s\\n "$count"

My main interest is in checking if the execute bit is not set for all. That is not just for the current user.

CodePudding user response：

The following counts directories without executable bit for current user:

find . -mindepth 1 -maxdepth 1 -type d '!' -executable -printf . | wc -c

The following counts directories that have 0 executable bits:

find . -mindepth 1 -maxdepth 1 -type d '!' -perm /111 -printf . | wc -c

The following counts directories that have 2 or less executable bits set:

find . -mindepth 1 -maxdepth 1 -type d '!' -perm -111 -printf . | wc -c

Parts from man find:

   -perm /mode                                                                                                                         
          Any  of  the permission bits mode are set for the file.

  -perm -mode
          All  of the permission bits mode are set for the file.

   -executable                                                                                                                         
          Matches files which are executable and directories which are
          searchable (in a file name resolution sense) by the  current
          user.

CodePudding user response：

This oneliner should do. Add a | wc -l at the end if you want to just count the number:

ls | while read dir; do stat --printf="%A %n\n" "$dir"  | grep '^d' | grep -v '^d..x......' | grep -v '^d.....x...' | grep -v '^d........x'; done

And the result:

/tmp % mkdir b
/tmp % mkdir c
/tmp % chmod 600 a
/tmp % chmod 060 a
/tmp % chmod 600 a
/tmp % chmod 060 b
/tmp % chmod 006 c
/tmp % ls | while read dir; do stat --printf="%A %n\n" "$dir"  | grep '^d' | grep -v '^d..x......' | grep -v '^d.....x...' | grep -v '^d........x'; done
drw------- a
d---rw---- b
d------rw- c