Linux 当名称不包含某些单词时删除文件
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remove files when name does NOT contain some words
提问by DocWiki
I am using Linux and intend to remove some files using shell.
我正在使用 Linux 并打算使用 shell 删除一些文件。
I have some files in my folder, some filenames contain the word "good", others don't. For example:
我的文件夹中有一些文件，一些文件名包含“好”这个词，其他的则没有。例如：
ssgood.wmv
ssbad.wmv
goodboy.wmv
cuteboy.wmv
I want to remove the files that does NOT contain "good" in the name, so the remaining files are:
我想删除名称中不包含“good”的文件，所以剩下的文件是：
ssgood.wmv
goodboy.wmv
How to do that using rmin shell? I try to use
如何rm在shell中使用它？我尝试使用
rm -f *[!good].*
but it doesn't work.
但它不起作用。
Thanks a lot!
非常感谢！
采纳答案by EdoDodo
This command should do what you you need:
这个命令应该做你需要的：
ls -1 | grep -v 'good' | xargs rm -f
It will probably run faster than other commands, since it does not involve the use of a regex (which is slow, and unnecessary for such a simple operation).
它可能会比其他命令运行得更快，因为它不涉及使用正则表达式（这很慢，对于这样简单的操作来说是不必要的）。
回答by T.J. Crowder
You can use findwith the -notoperator:
您可以find与-not运算符一起使用：
find . -not -iname "*good*" -a -not -name "." -exec rm {} \;
I've used -execto call rmthere, but I wonder if it does, see below.findhas a built-in delete action
我曾经-exec在rm那里打电话，但我想知道它是否，见下文。find有内置的删除操作
But verycareful with that. Note in the above I've had to put an -a -not -name "."clause in, because otherwise it matched ., the current directory. So I'd test thoroughly with -printbefore putting in the -exec rm {} \;bit!
但对此非常小心。请注意，在上面我不得不放入一个-a -not -name "."子句，否则它会匹配.当前目录。所以我会-print在投入之前彻底测试-exec rm {} \;！
Update: Yup, I've never used it, but there is indeed a -deleteaction. So:
更新：是的，我从来没有用过它，但确实有一个-delete动作。所以：
find . -not -iname "*good*" -a -not -name "." -delete
Again, be careful and double-check you're not matching more than you want to match first.
同样，要小心并仔细检查您匹配的次数没有超过您想要匹配的次数。
回答by glenn Hymanman
With bash, you can get "negative" matchingvia the extglobshell option:
使用 bash，您可以通过shell 选项获得“否定”匹配extglob：
shopt -s extglob
rm !(*good*)
