如何在没有退出程序的情况下在bash中退出if语句？

Question

重写这个问题以避免更多的downvotes,因为我删除它为时已晚:

我正在编写一个脚本,要求用户确认并在sourcing其他脚本之前.为了简化代码,想象一下可能有两个脚本sourced,但我希望用户可以是sourcenone或者只是其中一个脚本 - 而不是两者.我试图使用if true source-script else exit不起作用的表单if声明,因为我将退出语句,但也作为一个整体的脚本,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.就像是:

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

第一:不要做这些.以其他方式构建您的程序.如果你向我们描述了为什么你认为你需要这种行为,我们可能会告诉你如何实现它.

回答这个问题:如果你把你的块包裹在一个循环中,你可以用来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,您可以使用函数,并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shell中(尽管这会阻止它执行其他操作,例如影响子shell外部代码的变量赋值):

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