我有一個for循環遞歸搜索目錄中的文件。從字符串路徑檢索日期
for FILE in $(find /home/mydir/ -name '*.txt' -or -name '*.zip')
它們具有以下格式：
/home/mydir/subdir1/subdir2/22280317.txt
我想提取離開.txt的作爲格式2017-03-28之日起6個號碼我已經用awk試過不過我有麻煩，因爲是我想忽略的兩位數字
這可能會解決你的問題
find -name "*.txt" -exec sh -c 'f=${0%.txt}; l6=${f: -6}; date -d "${l6: -2}-${l6:2:2}-${l6:0:2}" +"%Y-%m-%d" ' {} \;
測試結果
# 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}-${l6:0:2}" +"%Y-%m-%d" ' {} \;
2017-03-28
2017-03-17
1998-03-17
在情況下，如果你想與日期，那麼
[[email protected] tmp]$ 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表達式：
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/'
假設該文件是相同的格式（22280317.txt）命名爲永諾，這是我的解決方案：
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