bash 大于给定数字的数字的 Grep 行
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Grep lines for numbers greater than given number
提问by user2150250
I'm trying to grep for lines in the first field of an output that are greater than a given number. In this instance, that number is 755. Ultimately, what I'm doing is trying to list every file with privileges greater than (and not equal to) 755by using stat -c '%a %n' *and then pipe to some grep'ing (or possibly sed'ing?) to obtain this final list. Any ideas how this could best be accomplished?
我正在尝试 grep 输出的第一个字段中大于给定数字的行。在这种情况下，该数字是755。最终，我正在做的是尝试755通过使用stat -c '%a %n' *然后管道到一些 grep'ing（或可能 sed'ing？）来列出权限大于（且不等于）的每个文件以获得这个最终列表。任何想法如何最好地完成？
回答by Kent
try this:
尝试这个：
stat -c '%a %n' *|awk '>755'
if you just want the filename in your final output, skip the privilege numbers, you could:
如果您只想要最终输出中的文件名，请跳过权限编号，您可以：
stat -c '%a %n' *|awk '>755{print }'
EDIT
编辑
actually you could do the chmodwithin awk. but you should make sure the user execute the awk line has the permission to change those files.
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 '>755{system("chmod 755 ")}'
again, assume the filename has no spaces.
再次假设文件名没有空格。
回答by Carl Norum
I'd use awk(1):
我会用awk(1)：
stat -c '%a %n' * | awk ' > 755'
The awkpattern matches lines where the first field is greater then 755. You could add an action if you want to print a subset of a line or something different, too (See @Kent's answer).
该awk模式匹配第一个字段大于 755 的行。如果您想打印一行的子集或其他内容，也可以添加一个操作（请参阅@Kent 的回答）。
回答by Lev Levitsky
Neither grepnor sedare good at arithmetics. awkcould help (unfortunately I don't know it). But note that findcan be handy here, too:
既不grep也不sed擅长算术。awk可以提供帮助（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).
So what could work for you is:
那么对你有用的是：
find . -perm -755 -printf '%m %p\n'
Just remove the -printfpart if you only need filenames.
-printf如果您只需要文件名，只需删除该部分。
