在bash间接变量赋值 [英] Indirect variable assignment in bash

问题描述

似乎在bash中做间接变量设置推荐的方法是使用评估：

  VAR = X; VAL = foo的
EVAL是$ var = $ VAL
回声$ X＃ - ＆GT;富
 

问题就是平常的那个与<$​​ C $ C>评估：

  VAR = X; VAL = $ 1\\ n'pwd
EVAL是$ var = $＃VAL坏输出这里
 

（而且由于它在很多地方推荐的，我不知道，因为这有多少脚本是脆弱的......）

在任何情况下，使用明显的溶液（逃跑）的报价并没有真正的工作：

  VAR = X; VAL = 1 \\$'\\ n'pwd \\
EVAL是$ var = \\$ VAL \\＃失败，上述
 

的事情是，bash有可变间接引用出炉，但我没有看到任何这样的方式做间接分配（以 $ {富！}） - - 是否有任何理智的方式来做到这一点。

？

有关记录，我没有找到一个解决办法，但这不是东西，我会考虑理智......

 评估是$ var ='$ {// VAL \\'/ \\'\\\\'\\\\'}'
 

解决方案

主要的一点是，推荐的方式来做到这一点是：

 评估是$ var = \\ $ VAL
 

在RHS间接完成了。由于评估在相同的环境中使用，将有 $ VAL 的约束，因此推迟它的工作原理，既然现在它只是一个变量（与已知的名字），没有任何问题与引用。

Seems that the recommended way of doing indirect variable setting in bash is to use eval:

var=x; val=foo
eval $var=$val
echo $x  # --> foo

The problem is the usual one with eval:

var=x; val=1$'\n'pwd
eval $var=$val  # bad output here

(and since it is recommended in many places, I wonder just how many scripts are vulnerable because of this...)

In any case, the obvious solution of using (escaped) quotes doesn't really work:

var=x; val=1\"$'\n'pwd\"
eval $var=\"$val\"  # fail with the above

The thing is that bash has indirect variable reference baked in (with ${!foo}), but I don't see any such way to do indirect assignment -- is there any sane way to do this?

For the record, I did find a solution, but this is not something that I'd consider "sane"...:

eval "$var='"${val//\'/\'\"\'\"\'}"'"

解决方案

The main point is that the recommended way to do this is:

eval "$var=\$val"

with the RHS done indirectly too. Since eval is used in the same environment, it will have $val bound, so deferring it works, and since now it's just a variable (with a known name), there are no issues with quoting.

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