#!/bin/bash
for i in $(ls $1); do
echo -n $i | sed 's/.dat//g'
grep '<Overall>' $i | sed 's/<Overall>//g'
awk 'BEGIN{sum=0} {sum+=$1} END{print sum/NR}'
sed -re 's/([0-9]+\.[0-9]{2})[0-9]+/\1/g'
echo 1
done | sort -nrk2
此脚本应返回平均整体评分。我找不到错误,因为我没有得到任何输出。
答案 0 :(得分:0)
未经测试:
gawk -F'>' '
BEGINFILE {sum = n = 0}
$1 == "<Overall" {sum += $2; n++}
ENDFILE {print FILENAME, (n == 0) ? "n/a" : sum/n}
' *.dat