我这就是我想要得到的结果(除了工作)
find ./ -mindepth 1 -type f -mtime +60 -print0 | xargs -0 egrep -vZ 'vvv|iii'
我究竟做错了什么?
$ ll
total 0
-rw-rw-r-- 1 yyy yyy 0 Sep 18 10:36 iii.txt
-rw-rw-r-- 1 yyy yyy 0 Aug 29 10:35 old1.txt
-rw-rw-r-- 1 yyy yyy 0 Aug 29 10:35 old2.txt
-rw-rw-r-- 1 yyy yyy 0 Aug 29 10:35 old3.txt
-rw-rw-r-- 1 yyy yyy 0 Nov 16 09:36 vvv.txt
-rw-rw-r-- 1 yyy yyy 0 Nov 5 09:41 young.txt
$ find ./ -mindepth 1 -type f -mtime +60 -print0 | xargs -0 egrep -viZ 'vvv|iii'
$ find ./ -mindepth 1 -type f -mtime +60 -print0 | xargs -0 egrep -vilZ 'vvv|iii'
$ find ./ -mindepth 1 -type f -mtime +60 -print0
./old3.txt./old1.txt./old2.txt$ find ./ -mindepth 1 -type f -mtime +60 -print0 | xargs -0 egrep 'old'
$ find ./ -mindepth 1 -type f -mtime +60 -print0 | xargs -0 grep 'old'
$ find ./ -mindepth 1 -type f -mtime +60 -print0 | xargs -0 grep 'o'
$ find ./ -mindepth 1 -type f -mtime +60 -print0 | xargs -0 grep '.*o.*'
$ find ./ -mindepth 1 -type f -mtime +60 | xargs egrep 'o'
$ find ./ -mindepth 1 -type f -mtime +60 | xargs egrep '.*o.*'
$ find ./ -mindepth 1 -type f -mtime +60
./old3.txt
./old1.txt
./old2.txt
$ find ./ -mindepth 1 -type f -mtime +60 | grep 'o'
./old3.txt
./old1.txt
./old2.txt
$ find ./ -mindepth 1 -type f -mtime +60 | xargs grep 'o'
$ find ./ -mindepth 1 -type f -mtime +60 -print | xargs grep 'o'
$ find . -name "*.txt" | xargs grep "old"
$ find . -name "*.txt"
./old3.txt
./vvv.txt
./iii.txt
./old1.txt
./old2.txt
./young.txt
$ find ./ | grep 'o'
./old3.txt
./old1.txt
./old2.txt
./young.txt
$ find ./ | xargs grep 'o'
$
我需要 grep 因为排除列表最终将来自一个文件,所以仅使用 find 进行过滤还不够。我也希望 grep 返回一个NUL
终止列表。我将把这个结果通过管道传递给其他东西,所以我不知道 find 选项是否-exec
合适。
我看过的东西:
- https://stackoverflow.com/questions/1362615/how-to-find-files-containing-a-string-using-egrep
- http://www.unixmantra.com/2013/12/xargs-all-in-one-tutorial-guide.html
$ bash -version
GNU bash, version 3.2.25(1)-release (x86_64-redhat-linux-gnu)
Copyright (C) 2005 Free Software Foundation, Inc.
$ cat /proc/version
Linux version 2.6.18-371.8.1.0.1.el5 ([email protected]) (gcc version 4.1.2 20080704 (Red Hat 4.1.2-54)) #1 SMP Thu Apr 24 13:43:12 PDT 2014
免责声明:我没有很多 Linux 或 shell 经验。
答案1
看起来你想grep
在文件名上使用 buf 如果你这样做:
find ./ -mindepth 1 -type f -mtime +60 -print0 | xargs -0 egrep -vZ 'vvv|iii'
实际上xargs
显示了find
作为egrep
.
您应该如何处理 NUL 终止的输入(来自-print0
)
find ./ -mindepth 1 -type f -mtime +60 -print0 | xargs -0 grep -EvzZ 'vvv|iii'
(egrep
已弃用,这就是我将其更改为的原因grep -E
)
从man grep
:
-z, --null-data
Treat the input as a set of lines, each terminated by a zero
byte (the ASCII NUL character) instead of a newline. Like the
-Z or --null option, this option can be used with commands like
sort -z to process arbitrary file names.
-Z, --null
Output a zero byte (the ASCII NUL character) instead of the
character that normally follows a file name.
所以你需要-z
两者-Z