bash 查找不在列表中的文件
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find files not in a list
提问by rurouni
I have a list of files in file.lst.
Now I want to find all files in a directory dirwhich are older than 7 days, except those in the file.lstfile. How can I either modify the find command or remove all entries in file.lstfrom the result?
我有一个file.lst. 现在，我想在目录中查找dir超过 7 天的所有文件，file.lst文件中的文件除外。如何修改 find 命令或file.lst从结果中删除所有条目？
Example:
例子：
file.lst:
file.lst：
a
b
c
Execute:
执行：
find -mtime +7 -print > found.lst
found.lst:
found.lst：
a
d
e
so what I expect is:
所以我期望的是：
d
e
回答by dogbane
Pipe your findcommand through grep -Fxvf:
find通过管道传输您的命令grep -Fxvf：
find -mtime +7 -print | grep -Fxvf file.lst
What the flags mean:
标志的含义：
-F, --fixed-strings
Interpret PATTERN as a list of fixed strings, separated by newlines, any of which is to be matched.
-x, --line-regexp
Select only those matches that exactly match the whole line.
-v, --invert-match
Invert the sense of matching, to select non-matching lines.
-f FILE, --file=FILE
Obtain patterns from FILE, one per line. The empty file contains zero patterns, and therefore matches nothing.
回答by Fredrik Pihl
Pipe the find-command to grepusing the -vand -fswitches
通过管道将 find-command 用于grep使用-v和-f开关
find -mtime +7 -print | grep -vf file.lst > found.lst
grep options:
grep选项：
-v : invert the match
-f file: - obtains patterns from FILE, one per line
example:
例子：
$ ls
a b c d file.lst
$ cat file.lst
a$
b$
c$
$ find . | grep -vf file.lst
.
./file.lst
./d
