ZSHfor循环数组变量问题-ZSHforlooparrayvariableissue

31  

It's actually much simpler than that:

它实际上比这简单得多：

lw=('plugin1' 'plugin2' 'plugin3')

for i in $lw; do
  . ~/Library/Rogall/plugins/$i/lw.prg end
done

In summary:

综上所述：

• Assign to "foo", not "$foo" (the shell would try to expand $foo and assign to whatever it expands to; typically not useful)
• 分配给“foo”，而不是“$ foo”（shell将尝试扩展$ foo并分配给它扩展到的任何内容;通常没有用）
• Use the loop variable directly; it contains the array value rather than the index
• 直接使用循环变量;它包含数组值而不是索引

15  

Why bother using the array? This can be done in portable sh very easily:

为什么要使用这个阵列呢？这可以很容易地在便携式中完成：

lw='plugin1 plugin2 plugin3'

for i in $lw;
  do . ~/Library/Rogall/plugins/$i/lw.prg end
done

Note that for this to work in zsh, you need to make zsh do the right thing with: set -o shwordsplit

请注意，要使其在zsh中工作，您需要使用以下命令使zsh做正确的事：set -o shwordsplit