Linux – Show the matching count of all files that contain a word

You're making two different mistakes that play together. First, you need to quote the pattern sent to grep, otherwise the shell will expand it first. (That's where the error message comes from.) Second, grep does not accept shell globs, it wants a regex.

zsh being what it is, you may want to say

$ ls **/*.{cpp,h} | wc -l

instead, using a zsh-style recursive glob. If you want to use the other one, it's

$ ls -R | egrep '\.(cpp|h)$' | wc -l

First you would use the find command in the terminal.

find . -type f -name "*.what"

That will list all files on the system from the current directory "." matching "type: file" and name "*.what".

So you can incorporate that into a bash script, like so:

Edit

Here you go, this does what you want I think.

#!/bin/bash

src=${1:-"."}
ext=${2:-"what"}

for dir in `find ${src} -type f -name "*.${ext}"`; do
    dir=`echo ${dir} | awk 'BEGIN{FS=OFS="/"}{$NF=""}{print}'`
    echo ${dir} "has" `ls -l ${dir} | awk '!NR=1 && !/^d/ && !/*.what/ {print $NF}' | wc -l` "file(s)"
done

That will output the number of files in any directories that contain *.what (recursively). The number if files excludes directories !/^d/ and the *.what file !/*.what/.

That should get you there. Works on my system at least, assuming I understand the question.