Identify sub-directories that do not contain a specific string in a specific file

Question

I have a directory named dir1 which has about 800 sub-directories named disp-001, disp-002, ... disp-800. I need to find sub-directories that

• either do not contain a file stdout or
• if they do, that file does not contain a particular string str1.

Identifying sub-directories that do not contain the file is answered in another question

$ find . -type d \! -exec test -e '{}/stdout' \; -print

However if I try to include grep in the above command, it does not work

 $ find . -type d \! -exec test -e 'grep str1 {}/stdout' \; -print

How can I include the string search to return me the directory of interest?

Answer 1

You can adapt any of the solutions there, e.g.

• use ( -exec or -exec ) with slm's or patrick's solutions (the second exec is executed only if the first one returns false, -print is executed only if one of them returns true):

  find . -type d \( ! -exec test -f '{}/stdout' \; -o ! -exec grep -q str1 '{}/stdout' \; \) -print
  

  or even shorter as suggested by Costas:

  find . -type d \! -exec grep -q 'str1' {}/stdout 2>/dev/null \; -print
  

• use a condition with terdon's solution:

  for d in **/
  do
    if [[ ! -f "$d"stdout ]] then
      printf '%s\n' "$d"
    else
      grep -q str1 "$d"stdout || printf '%s\n' "$d"
    fi
  done
  

• or, with zsh:

  print -rl **/*(/e_'[[ ! -f $REPLY/stdout ]] || ! grep -q str1 $REPLY/stdout'_)
  

Answer 2

To get that list just use grep:

grep -L str1 dir-*/stdout

Where:

• -L gives only the file without a match.
• str1 is the string you want to search.
• And if your files are in the same depth you can use simple wildcards.
  • If not, use the -r flag of grep to search recusively in the directories.

---

To continue that and work on that list you could pipe it nullbyte-delimited (greps -Z) to a while loop:

grep -LZ str1 dir-*/stdout | while IFS= read -r -d '' f; do
  echo "${f%%/*}" # gives the directory name
done