无法在bash中的括号内访问变量 [英] Can't access variables inside parenthesis in bash

问题描述

在bash中，如果我跑步

In bash, if I run

(foo=14)

然后稍后在我的bash脚本中尝试引用该变量:

And then try to reference that variable later on in my bash script:

echo "${foo}"

我什么都没得到.如何让bash像我需要的那样存储此变量?

I don't get anything. How can I make bash store this variable the way I need it to?

具体来说，我在if语句中使用此代码并检查退出代码，例如:

Specifically, I am using this in an if statement and checking the exit code, something kind of like:

if (bar="$(foo=14;echo "${foo}"|tr '1' 'a' 2>&1)")
then
    echo "Setting "'$bar'" was a success. It is ${bar}"
else
    echo "Setting "'$bar'" failed with a nonzero exit code."
fi

推荐答案

括号中的命令，例如()在子外壳中执行.子外壳中的任何分配都不会存在于该子外壳之外.

Commands enclosed in parenthesis e.g. () are executed in a sub-shell. Any assignment in a sub-shell will not exist outside that sub-shell.

foo=14
bar=$(echo $foo | tr '1' 'a' )
if [[ $? -eq 0 ]]
then
    echo "Setting "'$bar'" was a success. It is ${bar}"
else
    echo "Setting "'$bar'" failed with a nonzero exit code."
fi

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