bash - 如何在不退出程序的情况下退出bash中的if语句？

Question

重写这个问题以避免更多的反对，因为我删除它为时已晚:

我正在编写一个脚本，在 sourcing 一些其他脚本之前要求用户确认。为了简化代码，假设有两个脚本可能是sourced，但我希望用户要么source 没有一个脚本，要么只使用其中一个脚本——而不是两者。我试图使用形式为 if true source-script else exit 的语句，它不起作用，因为我会退出 if 语句，但脚本也是一个整体，并且没有机会进行必要的清理。最初，我的脚本看起来像这样:

echo "This script might do something terrible to your computer."
read -p "Do you wish to continue? (y/[n]) " -n 1;
echo
if ! [[ $REPLY =~ ^[Yy]$ ]]
then
    source "terrible_script.sh"
    # want some way to ensure that we don't prompt the user about the next script
    # don't want to just exit if the response is 'n' because we have to do cleanup
fi

echo "This script might do something really good to your computer."
read -p "Do you wish to continue? (y/[n]) " -n 1;
echo
if ! [[ $REPLY =~ ^[Yy]$ ]]
then
    source "good_script.sh"
fi

# do cleanup here
# regardless of whether or not any scripts were sourced

@charles-duffy 提供了答案 - 只需将提示包装在一个函数中。比如:

function badscript() {
    echo "This script might do something terrible to your computer."
    read -p "Do you wish to continue? (y/[n]) " -n 1;
    echo
    if ! [[ $REPLY =~ ^[Yy]$ ]]
    then
        source "terrible_script.sh"
        return 0
    fi
}

function goodscript() {
    echo "This script might do something really good to your computer."
    read -p "Do you wish to continue? (y/[n]) " -n 1;
    echo
    if ! [[ $REPLY =~ ^[Yy]$ ]]
    then
        source "good_script.sh"
    fi
}

if ! badscript
then
    goodscript
fi

# cleanup code here

Answer 1

首先:不要这样做。以其他方式构建您的程序。如果您向我们描述了为什么您认为您需要这种行为，我们可能会告诉您如何以其他方式实现它。

---

回到问题:如果您将 block 包装在一个循环中，您可以使用 break 提前退出:

for _ in once; do
  if true; then
    echo "in the loop"
    break
    echo "not reached"
  fi
done
echo "this is reached"

---

或者，你可以使用一个函数，然后return提前退出:

myfunc() {
  if true; then
    echo "in the loop"
    return
  fi
  echo "unreached"
}
myfunc
echo "this is reached"

---

或者，您可以将循环包装在子外壳中(尽管这会阻止它执行其他操作，例如影响子外壳外部代码的变量赋值):

(if true; then
   echo "in the block"
   exit
   echo "unreached"
 fi)
echo "this is reached."