Unix shell 代码。 如何计算文本文件中每一行的字母数

Question

I have the following code in Linux shell file: I need to replace the "..." with the number of letters of the coresponding line.

#!/bin/sh
echo "Filename is: $1\n"
nr_lines=$(wc -l <$1)
echo "Number of lines in files is: $nr_lines\n"

for line in $(seq 1 $nr_lines);
do
    echo "Line $line has  ... letters"


done 

Answer 1

The general pattern for iterating over all lines of a file is something like:

i=0; while read line; do
    printf "Line $((++i)) has %d letters\n" \
      "$(echo "$line" | tr -dc a-zA-Z | wc -c)";
done < input

but using while read...; do... done < input while read...; do... done < input in a shell is often better done with awk:

awk '{gsub("[^a-zA-Z]", ""); printf "Line %d has %d letters\n", NR, length}' input