BASH 使用双变量 $ - 错误替换
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BASH using double variable $ - bad substitution
提问by adiwhy
From my code below, how to make the value of 'zz' become 500 after replacing 'critical_' with x on variable 'yy'
从我下面的代码中，如何在变量 'yy' 上用 x 替换 'critical_' 后使 'zz' 的值变为 500
xab123=500
yy="critical_ab123"
zz=${"${yy//critical_/x}"}
echo $zz
instead the result, there is an error:
相反，结果，有一个错误：
line 8: ${"${yy//critical_/x}"}: bad substitution
thanks adi
谢谢阿迪
回答by anubhava
May be like this:
可能是这样的：
xab123=500
yy="critical_ab123"
zz="${yy//critical_/x}"
echo ${!zz}
500
回答by Stephen Quan
An interesting usage is when you call a bash function, you can use indirection on the parameters passed in. Then, you can nest calls to your indirection function in a nested manner using command substitution.
一个有趣的用法是，当您调用 bash 函数时，您可以对传入的参数使用间接。然后，您可以使用命令替换以嵌套方式嵌套对间接函数的调用。
deref() { echo "${!1}"; }
aa="bb"
bb="cc"
cc="hello"
echo "$(deref aa)" # bb
echo "$(deref "$(deref aa)")" # cc
echo "$(deref "$(deref "$(deref aa)")")" # hello
Here's derefused to solve the OP's problem:
这里deref用于解决OP的问题：
deref() { echo "${!1}"; }
xab123="500"
yy="critical_ab123"
zz="$(deref "${yy//critical_/x}")"
echo "$zz" # Outputs: 500
Applied edits based on @charles-duffy comments:
基于@charles-duffy 评论的应用编辑：
- Disclaimer: reader beware, there are performance impacts to the command substitution used in this approach (FIFO creation, fork() of a subshell, read() and wait().
- Quotes were added to protect against lossy expansion, i.e.
echo "$zz"is better thanecho $zz - Use POSIX-compliant function declaration syntax, i.e. replaced
function deref { echo "${!1}" ; }withderef() { echo "${!1}" ; } - Corrected quoting issue for each quoting context
- 免责声明：读者请注意，此方法中使用的命令替换会影响性能（FIFO 创建、子 shell 的 fork()、read() 和 wait()）。
- 添加了引号以防止有损扩展，即
echo "$zz"优于echo $zz - 使用符合 POSIX 的函数声明语法，即替换
function deref { echo "${!1}" ; }为deref() { echo "${!1}" ; } - 更正每个引用上下文的引用问题
