在 zsh 中回显全局变量 [英] Echoing the global variables in zsh

问题描述

我注意到检查 PATH、GOPATH 等全局变量是很常见的任务.这就是为什么我想写一个小函数而不是输入很多字母

I've noticed that it's usual task to check global variables like PATH, GOPATH and etc. That's why I want to write small function so instead of typing a lot of letters

echo $PATH

我只能打字

e PATH

函数本身应该很简单

function e() {
  echo $($1)   # it produces the error "command not found"
}

但是问题是如何替换一个变量来获取PATH的内容?

But the problem is how to substitute a variable to get the content of PATH?

附言我正在使用 zsh

P.S. I'm using zsh

推荐答案

处理此问题的传统 (POSIX) 符号使用 eval 命令，许多人会警告您:

The traditional (POSIX) notation to handle this uses the eval command, which many will warn you against:

e() {
  eval echo \"\$$1\"
}

然而，在 bash 中，您可以使用变量间接:

In bash, however, you can use variable indirection:

function e() {
  printf '%s\n' "${!1}"
}

在 zsh 中，您在我最初的回答后添加了它作为标签，间接处理的处理方式不同:

And in zsh, which you added as a tag after my initial answer, indirection is handled differently:

function e() {
  printf '%s\n' "${(P)1}"
}

这使用了一个参数扩展标志，您可以在 man zshexpn 中阅读.

This uses a Parameter Expansion Flag which you can read about in man zshexpn.

   P   This forces the value of the parameter name to be interpreted as a
       further parameter name, whose value will be used where  appropriate.

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