如何计算所有文件行

Question

I also need the directory name to be outputs as well. What I was able to do is to output the total number of lines in all directories with directory name.

find . -name '*.c' | xargs wc -l | xargs -I{} dirname {} | xargs -I{} dirname {}

Answer 1

Here is a script:

#!/usr/bin/env bash

for dir in */; do (
    cd "$dir"
    count=$(find . -name '*.c' -print0 | xargs -0 grep '[;]$' | wc -l)
    echo -e "${count}\t${dir}"
) done

If you want numbers for each sub-directory:

#!/usr/bin/env bash

for dir in $(find . -type d); do (
    cd "$dir"
    count=$(find . -maxdepth 1 -name '*.c' -print0 | \
        xargs -0 grep '[;]$' | wc -l)
    echo -e "${count}\t${dir}"
) done

Using -maxdepth 1 makes sure the calculation is only done in the current directory, not its sub-directories. So each file is counted once.

Answer 2

I have jumbled up a mixture of bash commands mostly GNU -specific, make sure you have them, GNU grep and GNU Awk

find . -type f -print0 | xargs -0 grep -c ';$' | \
  awk -F":" '$NF>0{cmd="dirname "$1; while ( ( cmd | getline result ) > 0 ) {printf "%s\t%s\n",result,$2} close(cmd) }'

The idea is grep -c returns the pattern count in format, file-name:count , which I am passing it to GNU Awk to filter those files whose count is greater than zero and print the directory of the file containing it and the count itself.

As a fancy one-liner as they call it these days,

find . -type f -print0 | xargs -0 grep -c ';$' | awk -F":" '$NF>0{cmd="dirname "$1; while ( ( cmd | getline result ) > 0 ) {printf "%s\t%s\n",result,$2} close(cmd) }'