linux-shell下刪除當前文件夾中按時間排序的前N個文件夾

#!/bin/sh

ls -lrt| awk '{if(NR<4 && NR!=1) printf("rm -rf %s\n",$8)}' >rm2.sh

chmod 755 rm2.sh

./rm2.sh

rm rm2.sh

 

或

#!/bin/sh

for i in `ls -lthr | head -3 |grep -v 'total'`

do

 rm -rf  $i

done;

參考： https://blog.csdn.net/osyun/article/details/6650519