bash - Unix - 按文件名过滤查找结果

Question

在 unix bash 中，我得到的文件如下所示:

find . -regextype posix-egrep -iregex "./[0-9]{8}/exp/RESULT.out"| sort

./20150128/exp/RESULT.out
./20150210/exp/RESULT.out
./20150218/exp/RESULT.out
./20150224/exp/RESULT.out
./20150318/exp/RESULT.out
./20150327/exp/RESULT.out
./20150330/exp/RESULT.out

是否可以按日期 YYYYMMDD 日期字符串过滤此结果，以便我只获取指定日期之前并包括指定日期的文件？

例如对于 20150224，我只想得到

./20150128/exp/RESULT.out
./20150210/exp/RESULT.out
./20150218/exp/RESULT.out
./20150224/exp/RESULT.out

谢谢你的帮助

Answer 1

您可以使用此查找 命令:

export s='20150224'

find . -regextype posix-egrep -iregex "./[0-9]{8}/exp/RESULT.out" -exec \
     bash -c 'IFS=/ read -ra arr <<< "$1"; ((${arr[1]} <= s)) && echo "$1"' - '{}' \;
./20150128/exp/RESULT.out
./20150210/exp/RESULT.out
./20150218/exp/RESULT.out
./20150224/exp/RESULT.out