Unix shell 代碼。 如何計算文本文件中每一行的字母數

Question

我在 Linux shell 文件中有以下代碼：我需要將“...”替換為對應行的字母數。

#!/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

遍歷文件所有行的一般模式類似於：

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

但是while read...; do... done < input while read...; do... done < input通常最好使用 awk 完成：

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