当存储在变量中时，subshel​​l的括号不起作用 [英] Parentheses for subshell don't work when stored in a variable

问题描述

命令:

( echo 1 )

当我在命令行中输入它时，它工作正常，但是如果将其存储为变量并调用它，则会出现错误:

works fine when I input it in the command line but if I store it as a variable and call it, it gives the error:

(echo: command not found

代码:

input="( echo 1 )"
$input

为什么当我这样称呼括号时，它不以相同的方式评估括号并将其放入子壳中?

Why doesn't it evaluate the parentheses the same way and put it into a subshell when I call it this way?

推荐答案

这在 BashFAQ中进行了详细讨论. #50 .

无引号的扩展仅经历两个阶段的shell解析:字段拆分和全局扩展.因此，首先将( echo 1 )分为以下几个字段:(，echo，1和)；每个扩展为一个全局扩展(模拟，因为它们都不是全局扩展)；然后它们作为命令运行:(被调用，第一个参数echo，第二个参数1和第三个参数).

An unquoted expansion goes through only two stages of shell parsing: Field-splitting, and glob expansion. Thus, ( echo 1 ) is first split into fields: (, echo, 1, and ); each is expanded as a glob (moot, as none of them are glob expansions); and then they're run as a command: ( is invoked, with the first argument echo, the second argument 1, and the third argument ).

正确存储代码的方法是在函数中

The Right Way to store code is in a function:

# best-practices approach
input() ( echo 1; )
input

...或者，如果您想让人类读者更明确地看到，您真的想要一个子shell，并且不使用括号而不是出于错误或习惯而大括号:

...or, if you want to make it more explicit to a human reader that you really want a subshell and weren't using parens rather than braces by error or habit:

# same, but more explicit about intent
input() { (echo 1); }
input

...如果不可能的话，可以使用eval(但请注意 Bash常见问题#48 ):

...if not possible, one can use eval (but be wary of the caveats given in BashFAQ #48):

# avoid this approach if at all possible
input="( echo 1 )"
eval "$input"

---

如果要在字符串中构建命令的真正原因是要参数化其内容，请改用数组:

---

If the real reason you're building a command in a string is to parameterize its contents, use an array instead:

input_args=( 1 )                    # define an array
input() ( echo "${input_args[@]}" ) # use that array in a function (if needed)

# add things according to conditional logic as appropriate
if (( 2 > 1 )); then
  input_args+=( "possible argument here" )
fi

# call the function, or just use the array directly, such as: (echo "$(input_args[@]}" )
input

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