shell - 如何在shell函数中获得 "set -e"的效果和用处？

Question

set -e (或以 #!/bin/sh -e 开头的脚本)对于在出现问题时自动轰炸非常有用。它使我不必对每个可能失败的命令进行错误检查。

如何在函数中获得与 this 等效的值？

例如，我有以下脚本，该脚本在出现错误时立即退出并显示错误退出状态:

#!/bin/sh -e

echo "the following command could fail:"
false
echo "this is after the command that fails"

输出符合预期:

the following command could fail:

现在我想把它包装成一个函数:

#!/bin/sh -e

my_function() {
    echo "the following command could fail:"
    false
    echo "this is after the command that fails"
}

if ! my_function; then
    echo "dealing with the problem"
fi

echo "run this all the time regardless of the success of my_function"

预期输出:

the following command could fail:
dealing with the problem
run this all the time regardless of the success of my_function

实际输出:

the following output could fail:
this is after the command that fails
run this all the time regardless of the success of my_function

(即函数忽略 set -e )

这大概是预期的行为。我的问题是:如何获得 set -e的效果和用处在shell函数中？我希望能够设置一些东西，这样我就不必单独检查每个调用的错误，但是脚本会在遇到错误时停止。它应该尽可能地展开堆栈，直到我检查结果，或者如果我没有检查它就退出脚本本身。这是什么 set -e已经，除了它不嵌套。

我找到了 the same question在 Stack Overflow 之外询问但没有合适的答案。

Answer 1

来自 set -e 的文档:

When this option is on, if a simple command fails for any of the reasons listed in Consequences of Shell Errors or returns an exit status value > 0, and is not part of the compound list following a while, until, or if keyword, and is not a part of an AND or OR list, and is not a pipeline preceded by the ! reserved word, then the shell shall immediately exit.

在您的情况下， false是以 ! 开头的管道的一部分和 if 的一部分.因此，解决方案是重写您的代码，使其不是。

换句话说，这里的函数没有什么特别之处。尝试:

set -e
! { false; echo hi; }