我正在尝试在输出的第一个字段中查找大于给定数字的行。在这种情况下,该数字是755
。最终,我正在做的是尝试列出具有大于(不等于)权限的每个文件,755
方法是使用stat -c '%a %n' *
然后管道到一些 grep'ing(或者可能是 sed'ing?)以获得这个最终列表。有什么想法可以最好地完成吗?
问问题
48441 次
3 回答
32
尝试这个:
stat -c '%a %n' *|awk '$1>755'
如果您只想在最终输出中使用文件名,请跳过权限编号,您可以:
stat -c '%a %n' *|awk '$1>755{print $2}'
编辑
实际上你可以chmod
在 awk 内做。但是您应该确保执行 awk 行的用户有权更改这些文件。
stat -c '%a %n' *|awk '$1>755{system("chmod 755 "$2)}'
再次假设文件名没有空格。
于 2013-04-30T22:32:02.570 回答
7
我会使用awk(1)
:
stat -c '%a %n' * | awk '$1 > 755'
该awk
模式匹配第一个字段大于 755 的行。如果要打印行的子集或其他内容,也可以添加一个操作(参见@Kent 的答案)。
于 2013-04-30T22:32:34.693 回答
4
也不grep
擅长sed
算术。awk
可以帮助(不幸的是我不知道)。但请注意,这find
在这里也很方便:
-perm mode
File's permission bits are exactly mode (octal or symbolic).
Since an exact match is required, if you want to use this form
for symbolic modes, you may have to specify a rather complex
mode string. For example -perm g=w will only match files which
have mode 0020 (that is, ones for which group write permission
is the only permission set). It is more likely that you will
want to use the `/' or `-' forms, for example -perm -g=w, which
matches any file with group write permission. See the EXAMPLES
section for some illustrative examples.
-perm -mode
All of the permission bits mode are set for the file. Symbolic
modes are accepted in this form, and this is usually the way in
which would want to use them. You must specify `u', `g' or `o'
if you use a symbolic mode. See the EXAMPLES section for some
illustrative examples.
-perm /mode
Any of the permission bits mode are set for the file. Symbolic
modes are accepted in this form. You must specify `u', `g' or
`o' if you use a symbolic mode. See the EXAMPLES section for
some illustrative examples. If no permission bits in mode are
set, this test matches any file (the idea here is to be consis‐
tent with the behaviour of -perm -000).
所以对你有用的是:
find . -perm -755 -printf '%m %p\n'
-printf
如果您只需要文件名,只需删除该部分。
于 2013-04-30T22:38:37.503 回答