我有一个for循环递归搜索目录中的文件。从字符串路径检索日期
for FILE in $(find /home/mydir/ -name '*.txt' -or -name '*.zip')
它们具有以下格式：
/home/mydir/subdir1/subdir2/22280317.txt
我想提取离开.txt的作为格式2017-03-28之日起6个号码我已经用awk试过不过我有麻烦，因为是我想忽略的两位数字
这可能会解决你的问题
find -name "*.txt" -exec sh -c 'f=${0%.txt}; l6=${f: -6}; date -d "${l6: -2}-${l6:2:2}-${l6:0:2}" +"%Y-%m-%d" ' {} \;
测试结果
# create sample files
[[email protected] test]$ touch 10280317.txt
[[email protected] test]$ touch 10170317.txt
[[email protected] test]$ touch 10170398.txt
# files created
[[email protected] test]$ ls *.txt
10170317.txt 10170398.txt 10280317.txt
# output using date command which takes care of year
# remove .txt
# extract last 6 char
# input year-mon-date to date command
[[email protected] test]$ find -name "*.txt" -exec bash -c 'f=${0%.txt}; l6=${f: -6}; date -d "${l6: -2}-${l6:2:2}-${l6:0:2}" +"%Y-%m-%d" ' {} \;
2017-03-28
2017-03-17
1998-03-17
在情况下，如果你想与日期，那么
[[email protected] tmp]$ pwd
/tmp
[[email protected] tmp]$ find -name "*.txt" -exec bash -c 'f=${0%.txt}; l6=${f: -6}; echo $f $(date -d "${l6: -2}-${l6:2:2}-${l6:0:2}" +"%Y-%m-%d") ' {} \;
./tss/test/10280317 2017-03-28
./tss/test/10170317 2017-03-17
./tss/test/10170398 1998-03-17
With sed表达式：
find /home/mydir/ -name '*.txt' -or -name '*.zip' | sed -E 's/.*([0-9]{2})([0-9]{2})([0-9]{2})\.txt/20\3-\2-\1/'
假设该文件是相同的格式（22280317.txt）命名为永诺，这是我的解决方案：
awk -v FPAT="([0-9]{2})" 'BEGIN{ FS = "/"; OFS = "-" }{ x = substr($NF, 3, 6); if (patsplit(x, a)) print a[1], a[2], "20"a[3] }'
打印：
28-03-2017