这意味着什么?我该如何解决?

zsh compinit: insecure directories, run compaudit for list.
Ignore insecure directories and continue [y] or abort compinit [n]?

运行compaudit将返回如下结果:

There are insecure directories:
/usr/local/share/zsh/site-functions

当前回答

大多数答案都提供了解决方案,但没有提到为什么会出现这种警告。以下是ZSH公司的一段摘录:

For security reasons compinit also checks if the completion system would use files not owned by root or by the current user, or files in directories that are world- or group-writable or that are not owned by root or by the current user. If such files or directories are found, compinit will ask if the completion system should really be used. To avoid these tests and make all files found be used without asking, use the option -u, and to make compinit silently ignore all insecure files and directories use the option -i. This security check is skipped entirely when the -C option is given.

因此,解决方案意味着解决以下一个(或所有)问题:

将当前用户设置为所有目录/子目录/文件的所有者,原因如下: compaudit | xargs chown -R "$(whoami)" 删除组/others文件的写权限,原因如下: Compaudit | xargs chmod go-w

另一种方法是使用跳过这些检查

compinit -u

但我并不建议这样做,因为把问题藏在地毯下只能在短期内解决问题。

其他回答

大多数答案都提供了解决方案,但没有提到为什么会出现这种警告。以下是ZSH公司的一段摘录:

For security reasons compinit also checks if the completion system would use files not owned by root or by the current user, or files in directories that are world- or group-writable or that are not owned by root or by the current user. If such files or directories are found, compinit will ask if the completion system should really be used. To avoid these tests and make all files found be used without asking, use the option -u, and to make compinit silently ignore all insecure files and directories use the option -i. This security check is skipped entirely when the -C option is given.

因此,解决方案意味着解决以下一个(或所有)问题:

将当前用户设置为所有目录/子目录/文件的所有者,原因如下: compaudit | xargs chown -R "$(whoami)" 删除组/others文件的写权限,原因如下: Compaudit | xargs chmod go-w

另一种方法是使用跳过这些检查

compinit -u

但我并不建议这样做,因为把问题藏在地毯下只能在短期内解决问题。

我最近在卡特琳娜也收到了同样的警告。 一个简单的解决方法是将其放在.zshrc的顶部

ZSH_DISABLE_COMPFIX=true

注:此答案为2012年数据。


这为我解决了问题:

$ sudo chmod -R 755 /usr/local/share/zsh/site-functions

出处:zsh邮件列表上的一个帖子


编辑:正如@biocyberman在评论中指出的那样。您可能还需要更新站点函数的所有者:

$ sudo chown -R root:root /usr/local/share/zsh/site-functions

在我的机器(OSX 10.9)上,我不需要这样做,但YMMV。

EDIT2:在OSX 10.11上,只有这样才能工作:

$ sudo chmod -R 755 /usr/local/share/zsh
$ sudo chown -R root:staff /usr/local/share/zsh

另外,user:staff是OSX上正确的默认权限。

我通过做来解决它

sudo chown -R root:staff /usr/local/share/zsh

在我的例子中,share/中的其他目录也分配了“staff”组

接受的答案在macOs Sierra(10.12.1)上不适用。必须从/usr/local递归吗

cd /usr/local
sudo chown -R <your-username>:<your-group-name> *

注意:你可以用whoami获取你的用户名,用id -g获取你的组