相关疑难解决方法(0)

set -e(或以脚本开头#!/bin/sh -e)对于在出现问题时自动轰炸非常有用.它使我不必错误地检查可能失败的每个命令.

如何在函数内获得相应的内容？

例如,我有以下脚本在出错时立即退出并出现错误退出状态:

#!/bin/sh -e

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

输出如预期:

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"

预期产量:

the following command could fail: …

我试图在循环中调用一个函数，并在它抛出时优雅地处理和继续。

如果我省略|| handle_error它只会像人们期望的那样停止整个脚本。

如果我离开|| handle_error那里，它将foo is fine在错误发生后打印并且根本不会执行handle_error。这也是预期的行为，这就是它的工作方式。

#!/bin/bash

set -e

things=(foo bar)

function do_something {
  echo "param: $1"

  # just throw on first loop run
  # this statement is just a way to selectively throw
  # not part of a real use case scenario where the command(s)
  # may or may not throw
  if [[ $1 == "foo" ]]; then
    throw_error
  fi

  # this line should not be executed when $1 is …

注意:如果您正在寻找一种解决方法,因为set -e在函数中不起作用,请转到函数中的"set -e".这个问题是为什么它不能按预期工作.

当在GNU bash上运行以下版本时,版本4.1.5(1) - 释放:

  set -ex

  PS4=' ${FUNCNAME[0]}: $LINENO: '

  function f() {
      set -e
      false
      echo $@
  }

  ! f why does it display ? It should stop because of -e
  f

它显示

 : 11: f why does it display '?' It should stop because of -e
 f: 6: set -e
 f: 7: false
 f: 8: echo why does it display '?' It should stop because of -e
why …