count of zeros between the 1st and 14th digit by using grep [closed]

Answer 1

A (mostly) Bash solution using shell parameter expansion to get the first 14 characters:

file='500000000000922801808800000000000000000000922891863600000GS5'
count=$(echo "${file:0:13}" | grep --fixed-strings --only-matching 0 | wc -l)
if [[ $count -ne 12 ]]; then 
    mv "$file" "$drp_dir"
else
    mv "$file" "$dst_dir"
fi

You might also use head -c instead if you prefer.

Answer 2

Assuming that dst_dir and drp_dir are two output files you want to create, the following awk program will do that for you.

awk '
NF {
    data  = substr ($0, 1, 14)
    count = gsub (/0/, "", data)
    print > (count==12 ? "dst_dir" : "drp_dir")
}' file

Using NF we skip the blank lines. We create a subset of your line using substr function. gsub returns the number of substitution made, so we capture the return value to variable count.

Lastly, we test if the count is 12. If it is we write the record to dst_dir else we write the record to drp_dir.