如何递归获取目录下所有文件的总大小

Question

I am using this command to get the files less than 17MB:

hadoop fsck /admin_test -files | 
gawk '{if ($2 ~ /^[0-9]+$/ && $2 <= 17825792) print $1,$2;}'

How can I get the total size of all files less than 17MB?

Answer 1

gawk '
    $2 ~ /^[0-9]+$/ && $2 <= 17825792 {sum += $2; print $1, $2} 
    END {print "sum=", 0+sum}
'

Answer 2

what about using du with the --threshold=SIZE arg:

       -t, --threshold=SIZE
          exclude entries smaller than SIZE if positive, or entries greater than SIZE if negative

something like this:

du -sk --threshold=-17825792 /admin_test