shell – “set -e”在一个函数中

set -e(或以#！/ bin / sh -e开头的脚本)对于自动弹出来是非常有用的,如果有问题.它可以节省我错误检查每个可能失败的命令.

如何在函数中获得相当于此的内容？

例如,我有以下脚本立即退出错误退出状态：

#!/bin/sh -e

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

产出如预期的那样：

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"

预期产量：

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)

这可能是预期的行为.我的问题是：如何在shell函数中获得set -e的效果和有用性？我发现the same question在Stack Overflow外面询问,但没有合适的答案.

从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.

在你的情况下,虚假是管道之前的一部分！和if的一部分.所以解决方案是重写你的代码,使它不是.

换句话说,这里没有什么特别的功能.尝试：

set -e
! { false; echo hi; }

原文链接：https://www.f2er.com/bash/386243.html

shell – “set -e”在一个函数中

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