Combining greps to make script to count files in folder - linux

Question

I need some help combining elements of scripts to form a read output.
Basically I need to get the file name of a user for the folder structure listed below and using count the number of lines in the folder for that user with the file type *.ano
This is shown in the extract below, to note that the location on the filename is not always the same counting from the front.
/home/user/Drive-backup/2010 Backup/2010 Account/Jan/usernameneedtogrep/user.dir/4.txt
/home/user/Drive-backup/2011 Backup/2010 Account/Jan/usernameneedtogrep/user.dir/3.ano
/home/user/Drive-backup/2010 Backup/2010 Account/Jan/usernameneedtogrep/user.dir/4.ano
awk -F/ '{print $(NF-2)}'
This will give me the username I need but I also need to know how many non blank lines they are in that users folder for file type *.ano. I have the grep below that works but I dont know how to put it all together so it can output a file that makes sense.
grep -cv '^[[:space:]]*$' *.ano | awk -F: '{ s+=$2 } END { print s }'
Example output needed
UserA 500
UserB 2
UserC 20

Answer 1

find /home -name '*.ano' | awk -F/ '{print $(NF-2)}' | sort | uniq -c
That ought to give you the number of "*.ano" files per user given your awk is correct. I often use sort/uniq -c to count the number of instances of a string, in this case username, as opposed to 'wc -l' only counting input lines.
Enjoy.

Answer 2

Have a look at wc (word count).

Answer 3

To count the number of *.ano files in a directory you can use
find "$dir" -iname '*.ano' | wc -l
If you want to do that for all directories in some directory, you can just use a for loop:
for dir in * ; do
echo "user $dir"
find "$dir" -iname '*.ano' | wc -l
done

Answer 4

Execute the bash-script below from folder
/home/user/Drive-backup/2010 Backup/2010 Account/Jan
and it will report the number of non-blank lines per user.
#!/bin/bash
#save where we start
base=$(pwd)
# get all top-level dirs, skip '.'
D=$(find . \( -type d ! -name . -prune \))
for d in $D; do
cd $base
cd $d
# search for all files named *.ano and count blank lines
sum=$(find . -type f -name *.ano -exec grep -cv '^[[:space:]]*$' {} \; | awk '{sum+=$0}END{print sum}')
echo $d $sum
done

Answer 5

This might be what you want (untested): requires bash version 4 for associative arrays
declare -A count
cd /home/user/Drive-backup
for userdir in */*/*/*; do
username=${userdir##*/}
lines=$(grep -cv '^[[:space:]]$' $userdir/user.dir/*.ano | awk '{sum += $2} END {print sum}')
(( count[$username] += lines ))
done
for user in "${!count[#]}"; do
echo $user ${count[$user]}
done

Answer 6

Here's yet another way of doing it (on Mac OS X 10.6):
find -x "$PWD" -type f -iname "*.ano" -exec bash -c '
ar=( "${#%/*}" ) # perform a "dirname" command on every array item
printf "%s\000" "${ar[#]%/*}" # do a second "dirname" and add a null byte to every array item
' arg0 '{}' + | sort -uz |
while IFS="" read -r -d '' userDir; do
# to-do: customize output to get example output needed
echo "$userDir"
basename "$userDir"
find -x "${userDir}" -type f -iname "*.ano" -print0 |
xargs -0 -n 500 grep -hcv '^[[:space:]]*$' | awk '{ s+=$0 } END { print s }'
#xargs -0 -n 500 grep -cv '^[[:space:]]*$' | awk -F: '{ s+=$NF } END { print s }'
printf '%s\n' '----------'
done