regex - 删除名称长于x的目录

Question

我需要删除 Linux shell 上名称长度超过 4 个字符的目录。 但不计算子目录的长度。

例如:

/12345/..    <= Should be deleted
/123456/..   <= Should be deleted
/1234/12345  <= Should NOT be deleted
/1234/123456 <= Should NOT be deleted

更新: 明白了:

find -maxdepth 1 -regextype posix-egrep -type d -regex '.*[^/]{4}' -exec rm -rf {} +

Answer 1

要删除 bash 中所有包含 5 个或更多字符的目录，您可以执行以下操作:

rm -rf ?????*/

该表达式不是正则表达式，而是使用一组通配符来指定文件名或路径的 glob 模式。

基本上，如果您想要保留 4 个或更少字符的目录，则需要删除包含 5 个或更多字符的所有内容，因此使用 5 个 ? 和单个 *。 / 表示目录。

man bash

* :: Matches any string, including the null string. When the globstar shell option is enabled, and * is used in a pathname expansion context, two adjacent *s used as a single pattern will match all files and zero or more directories and subdirectories. If followed by a /, two adjacent *s will match only directories and subdirectories.

? :: Matches any single character.

$ find .
.
$ mkdir -p {1,12,123,1234,12345,123456}/{123,12345}
$ touch foobar
$ rm -rf ?????*/
$ find .
.
./123
./123/12345   <= subdirectory with 5 or more not deleted
./123/123
./foobar      <= the file is still here
./1234
./1234/12345
./1234/123
./12
./12/123
./12/12345
./1
./1/12345
./1/123