我试图在一组特定的目录中强制执行777的文件权限.我使用了“setfacl -m d：o :: rwx”并获得了正确的权限
$getfacl . # file: . # owner: blah # group: blah # flags: -s- user::rwx group::rwx other::rwx default:user::rwx default:group::rwx default:other::rwx
当我运行mkdir时,我得到了一个带有正确烫发的目录.
$mkdir test $ll -d test drwxrwsrwx+ 2 blah blah 4096 Oct 28 10:26 test
当我运行“mkdir -p”时,我获得与umask匹配的perms,而不是acl.
$mkdir -p test1 $ll -d test1 drwxrwsr-x+ 2 blah blah 4096 Oct 28 10:27 test1
有什么我想念的吗？
我相信这是正确的行为.看看信息mkdir：
`-p'
`--parents'
Make any missing parent directories for each argument,setting
their file permission bits to the umask modified by `u+wx'. Ignore
existing parent directories,and do not change their file
permission bits.
To set the file permission bits of any newly-created parent
directories to a value that includes `u+wx',you can set the umask
before invoking `mkdir'. For example,if the shell command
`(umask u=rwx,go=rx; mkdir -p P/Q)' creates the parent `P' it sets
the parent's permission bits to `u=rwx,go=rx'. To set a parent's
special mode bits as well,you can invoke `chmod' after `mkdir'.
*Note Directory Setuid and Setgid::,for how the set-user-ID and
set-group-ID bits of newly-created parent directories are
inherited.
因此,mkdir -p将获取umask值(由u rw修改)来创建不在树中的任何目录,如果您考虑如何解决已存在的父目录的权限问题,这种情况有何意义？
正如摘录所述,您可以在运行命令之前更改umask,尽管在创建所有内容后在父目录上运行递归chmod可能要容易得多.