print parent directories full path of find output

Question

what I want to do is to find all files based on some search query and get parents parent directory (../..) full path.

For example find . -name "__openerp__.py" and then for each file execute something along the lines of (cd ../..; pwd). Then pipe everything to uniq.

Answer 1

Assuming none of the file names contain newline characters:

find "$PWD" -name __openerp__.py  | awk -F/ -vOFS=/ 'NF-=2' | sort -u

Answer 2

Putting aside that there would be issues with file names that contain newlines, there are a couple of efficiency improvements I can think of:

1. Pass the $PWD directly to find, instead of determining it for each sub-shell. This will make find include the full path in its output.
2. Cut parent directory names off with rev and cut.

So I would suggest the following to get a list of all directories containing the desired file:

find "$PWD" -name '__openerp__.py' |
rev                                | 
cut -d/ -f3-                       |
rev

Note that find does not order its output, so it should be sorted before uniq'ing it. As sort already supports squeezing identical elements (-u switch), the original question can then be answered with:

find "$PWD" -name '__openerp__.py' |
rev                                | 
cut -d/ -f3-                       |
rev                                |
sort -u

Answer 3

If I understand correctly, you want the list of directories containing a subdirectory containing a file called __openerp__.py, without duplicates.

In zsh:

print -rl -- **/__openerp__.py(:h:h) | sort -u

or

a=(**/__openerp__.py(:h:h)); print -rl -- ${(u)a}

Answer 4

find . -name "__openerp__.py" -exec sh -c '(cd "{}"/../../; pwd)' \; | uniq