为什么像 [ $var == 'str' ] 这样的 zsh 代码作为命令运行良好，但作为脚本文件却出错?

问题描述

zsh_test.sh很简单，如下:

#!/usr/bin/env zsh

if [ $USER == 'root' ]; then
    echo "root"
else
    echo "not root"
fi

• 将以上代码复制粘贴到zsh shell中，运行良好.

  • Copy and paste the above codes into a zsh shell, it executed well.

    $ #!/usr/bin/env zsh
    $ if [ $USER == 'root' ]; then
    then>     echo "root"
    then> else
    else>     echo "not root"
    else> fi
    not root
    

  • 但是直接执行脚本文件zsh_test.sh，报错.

    $ ./zsh_test.sh
    ./zsh_test.sh:3: = not found
    

  推荐答案

  我现在明白出了什么问题:您是一个相当晦涩的 zsh 机制的受害者，该机制在 zshexpn 手册页和称为'='扩展.来自手册页:

  I now see what's wrong: You are the victim of a fairly obscure zsh mechanism, which is described in the zshexpn man page and is called '=' expansion. From the man-page:

  如果一个单词以不带引号的 `=' 开头并且设置了 EQUALS 选项，则该单词的其余部分将作为命令的名称.如果该名称存在命令，则该单词将替换为该命令的完整路径名.

  你可以用命令试试

  echo ==
  

  也输出此错误消息.例如，在我的平台上

  which also outputs this error message. For instance, on my platofm

  echo =ruby
  

  输出 /usr/bin/ruby，因为这是我安装 ruby​​ 的地方.如果您的 PATH 中有一个名为 = 的程序，则 == 将解析为该路径.

  outputs /usr/bin/ruby, because this is where I have ruby installed. If you would have in your PATH a program named =, the == would resolve to this path.

  虽然在 [ ... ] 中使用双 == 符号是不常见的，但此命令的 zsh 实现允许这样做，但您必须引用运算符，以避免 =-展开:

  While it is unusual to use a double == sign inside [ ... ], the zsh implementation of this command allows it, but you would have to quote the operator, to avoid =-expansion:

  if [ $USER '==' root ]; then
  

  另一种方法是使用 [[ ... ]] 代替.这不是一个命令，而是一个句法结构，它内部的扩展规则是不同的.因此

  An alternative would be to use [[ ... ]] instead. This is not a command, but a syntactic construct, and expansion rules are different inside it. Therefore

  if [[ $USER == root ]]; then
  

  也可以.

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