Question

在unix bash中，我得到的文件如下：

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的文件？

e.g。对于20150224，我只想得到

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

感谢您的帮助

Answer 1

通过

传递结果

awk -v FS=/ '$2<=20150224'

Answer 2

您可以使用此find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

Unix - 过滤按文件名查找结果

2 个答案: