linux - 如何根据最后一个单词PATH中的前两个字符对文件进行排序

Question

请指教如何根据最后一个单词 PATH 中的前两个字符对文件进行排序(使用 sort 命令)

   /dir/dir/dir/LAST_WORD_PATH

例子:

如果下面的路径是

      /Config/EMA/M5/Hatki Tekom/Customer Files/fSE-TMOLE.txt

所以最后一个单词PATH应该是

    fSE-TMOLE.txt

目标是对“last word PATH”中的前两个字符进行排序

然后前两个字符是(来自fSE-TMOLE.txt)

    fS

完整示例

排序前

     /Config/EMA/M5/Cmo Buria Mobile/dzs Bulia Mole.txt
     /Config/EMA/M5/Hatki Tekom/Customer Files/fAE-TMOLE.txt
     /Config/EMA/M5/Cmo Buria Mobile/dos Bulia Mole.txt
     /Config/EMA/M5/Cmo Bgaia Mobile/Fustoer files/Backup/Cos Buia Mole-2.txt
     /Config/EMA/M5/VFONE Roia/Cbdne.ro.txt
     /Config/EMA/M5/TEKOM SRJA/IELOM SRJA.txt
     /Config/EMA/M5/VP Moe/Bobile.txt

排序后

     /Config/EMA/M5/VP Moe/Bobile.txt
     /Config/EMA/M5/VFONE Roia/Cbdne.ro.txt
     /Config/EMA/M5/Cmo Bgaia Mobile/Fustoer files/Backup/Cos Buia Mole-2.txt
     /Config/EMA/M5/Cmo Buria Mobile/dos Bulia Mole.txt
     /Config/EMA/M5/Cmo Buria Mobile/dzs Bulia Mole.txt
     /Config/EMA/M5/Hatki Tekom/Customer Files/fAE-TMOLE.txt
     /Config/EMA/M5/TEKOM SRJA/IELOM SRJA.txt

Answer 1

这是一种方法:

#!/bin/bash
while read f
do
   b=$(basename "$f")   # Get basic name of file without path
   b2=${b:0:2}          # Extract first 2 letters
   echo $b2:$f          # Output for awk to read with colon separator
done < file | sort -f | awk -F: '{print $2}'