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
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
这两条线路已经为我固定了。
sudo chown -R _user_:root /usr/local/share/zsh
sudo chown -R _user_:root /usr/local/share/zsh/*
在macOS Sierra上,你需要运行: sudo chown -R $(whoami):staff /usr/local
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
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
这适用于我的Mac,因为High Sierra更新。
取消组写权限:
sudo chmod g-w /usr/local/share/zsh/site-functions
sudo chmod g-w /usr/local/share/zsh
最好将更改限制在zsh目录中。