如何获得“set -e"的效果和用处?在shell函数中? [英] How do I get the effect and usefulness of "set -e" inside a shell function?

问题描述

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.

set -e (or a script starting with #!/bin/sh -e) is extremely useful to automatically bomb out if there is a problem. It saves me having to error check every single command that might fail.

如何在函数中获得与 this 等效的值?

How do I get the equivalent of this inside a function?

例如，我有以下脚本，该脚本在出错时立即退出并显示错误退出状态:

For example, I have the following script that exits immediately on error with an error exit status:

#!/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:

Now I'd like to wrap this into a function:

#!/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

(即函数忽略了 set -e)

这大概是预期的行为.我的问题是:如何在 shell 函数中获得 set -e 的效果和用处?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 set -e 已经做的，除了它不嵌套.

This presumably is expected behaviour. My question is: how do I get the effect and usefulness of set -e inside a shell function? I'd like to be able to set something up such that I don't have to individually error check every call, but the script will stop on encountering an error. It should unwind the stack as far as is needed until I do check the result, or exit the script itself if I haven't checked it. This is what set -e does already, except it doesn't nest.

我发现 相同在 Stack Overflow 之外提出的问题，但没有合适的答案.

I've found the same question asked outside Stack Overflow but no suitable answer.

推荐答案

来自 set -e 的文档:

当这个选项打开时，如果一个简单的命令对于任何一个后果中列出的原因Shell 错误或返回退出状态值 > 0，并且不属于while 后面的复合列表，until 或 if 关键字，并且不是AND 或 OR 列表的一部分，并且不是以 ! 保留的管道字，那么壳应立即退出.

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 ! 和 前面的管道的一部分 if.因此，解决方案是重写您的代码，使其不是.

In your case, false is a part of a pipeline preceded by ! and a part of if. So the solution is to rewrite your code so that it isn't.

换句话说，这里的函数没有什么特别之处.试试:

In other words, there's nothing special about functions here. Try:

set -e
! { false; echo hi; }

这篇关于如何获得“set -e"的效果和用处?在shell函数中?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！