Handle parentheses in this xargs multi-command

Question

I suspect the following has been answered already but I don't know the terminology for the issue I'm having well enough to find an existing answer.

I'm working on a command to go through a list of files and output on each line the filename followed by the count of lines that start with P. I've gotten this far:

find -type f | xargs -I % sh -c '{ echo %; grep -P "^P \d+" % | wc -l; }  | tr "\n" ","; echo ""; '

_{(The actual find command is a bit more involved but short story is it finds about 11k files of interest in the directory tree below where I'm running this)}

This command is about 98% working for my purposes, but I discovered there is a small subset of files with parentheses in their names and I can't ignore them or permanently replace the parentheses with something else.

As a result I'm getting some cases like this:

sh: -c: line 0: syntax error near unexpected token `('

I know parentheses are a shell special character so for example if I was running grep directly on a single file with parentheses in the name I'd have to enclose the filename in single quotes or escape the parentheses. I tried swapping the quote types in my command (doubles outermost, singles inner) so I could put the '%' in the grep call in single quotes but that didn't help.

Is there a way to handle parentheses in the find -> xargs -> sh chain so they get handled correctly in the sh call?

Answer 1

Better not embed data (filenames) directly in code (the shell scriptlet). Instead pass the filename as an argument to the shell you have xargs run:

find -type f | xargs -I % \
  sh -c '{ echo "$1"; grep -c -P "^P \d+" "$1"; } | tr "\n" ","; echo' sh %

Also you should be able to use grep -c instead of grep | wc -l, it at least makes the command a bit shorter.

Answer 2

Since you omitted the . in find . -type f, I suppose your find is GNU find, then you can do:

find . -type f -printf %p, -exec grep -cP '^P \d' {} ';'

If the file paths don't contain : characters, you could also do (with GNU grep):

grep -rcP '^P \d' . | tr : ,

If they may contain : characters but don't contain newline characters, that can be worked around by replacing only the last : in the line with ,:

grep -rcP '^P \d' . | LC_ALL=C sed 's/\(.*\):/\1,/'

That approach can also be used with:

find ... -type f -exec grep -cHP '^P \d' {} + | ...

If you still need to use find, for instance because you have more selection criteria.

Answer 3

ilkkachu's answer looks like a fine improvement and is probably what you should do.

Adding, for information purposes, a lighter touch fix to show where your problem lies:

find -type f | xargs -I % sh -c '{ echo "%"; grep -P "^P \d+" "%" | wc -l; }  | tr "\n" ","; echo ""; '

Basically -- quote wrap the % that will be replaced.