在bash中对退出代码进行AND操作 [英] ANDing exit codes in bash

问题描述

我有一个bash脚本，对我的源代码进行三项检查，然后，如果所有命令都成功，则退出 exit 0 ，如果其中任何命令失败，则退出 exit 1 :

I have a bash script that runs three checks over my source code, and then exit 0 if all the commands succeeded, or exit 1 if any of them failed:

#!/bin/bash

test1 ./src/ --test-1=option
exit_1=$?

test2 ./src/ test-2-options
exit_2=$?

test3 ./src/ -t 3 -o options
exit_3=$?

# Exit with error if any of the above failed
[[ $exit_1 -eq 0 && $exit_2 -eq 0 && $exit_3 -eq 0 ]]
exit $?

此代码有效，但是感觉太冗长和冗长.有什么办法可以使它变得更好吗?具体来说，我不满意:

This code works, but it feels overly long and verbose. Is there some way this can be made nicer? Specifically I am not happy with:

• 必须运行命令，然后然后将退出代码分配给变量
• 必须使用 [[... ...]] ，然后在下一行收集其退出代码以退出
• 必须像在 [[$ $ -eq 0]] 中那样将变量显式比较为0，而不是将它们视为布尔值.

• Having to run the command, and then assign the exit code to a variable
• Having to use [[ ... ]], then collect its exit code on the next line to exit with
• Having to explicitly compare variables to 0, as in [[ $var -eq 0 ]], instead of treating them as booleans

理想情况下，最终结果将更具可读性，例如:

Ideally, the end result would be something more readable like:

exit_1=( test1 ./src/ --test-1=option )
exit_2=( test2 ./src/ test-2-options )
exit_3=( test3 ./src/ -t 3 -o options )

# Exit with error if any of the above failed
exit ( $exit_1 && $exit_2 && $exit_3 )

我考虑过的一些事情:

将错误代码放入一行中的变量中:

exit_1=$( test1 ./src/ --test-1=option )$?
exit_2=$( test2 ./src/ test-2-options )$?
exit_3=$( test3 ./src/ -t 3 -o options )$?

这有效，并且使它更短一些，但是我从未见过其他人使用过此功能.这是明智/理智的事情吗?这有什么问题吗?

This works, and makes this bit shorter, but I've never seen anyone else use this before. Is this a sensible/sane thing to do? Are there any issues with this?

只需运行测试，然后&&他们在一起:

test1 ./src/ --test-1=option && \
test2 ./src/ test-2-options && \
test3 ./src/ -t 3 -o options
status=$?

这不起作用，因为bash短路.如果 test1 失败，则test2和test3不运行，我希望它们全部运行.

This does not work, as bash short circuits. If test1 fails, test2 and test3 do not run, and I want them all to run.

检测错误并使用 ||退出退出

Detecing errors and exiting using || exit

[[ $exit_1 -eq 0 && $exit_2 -eq 0 && $exit_3 -eq 0 ]] || exit 1

这将保存一行尴尬的退出代码和变量，但是 exit 1 的重要位置现在位于该行的末尾，您可能会错过它.理想情况下，类似这样的方法会起作用:

This saves one line of awkward exit codes and variables, but the important bit of exit 1 is now right at the end of the line where you can miss it. Ideally, something like this would work:

exit [[ $exit_1 -eq 0 && $exit_2 -eq 0 && $exit_3 -eq 0 ]]

当然，这不是 的工作，因为 [[[]返回其输出而不是回显它.

Of course, this does not work, as [[ returns its output instead of echoing it.

exit $( [[ $exit_1 -eq 0 && $exit_2 -eq 0 && $exit_3 -eq 0 ]] ; echo $? )

确实有效，但看起来仍然很可怕

does work, but still seems like a horrid cludge

未明确处理布尔值退出代码

[[ $exit_1 && $exit_2 && $exit_3 ]]

这并没有您希望的那样.&&&最简单的方法存储在变量中的三个返回码一起是完整的 $ var -eq 0&&... .当然有更好的方法了吗?

This does not do what you would hope it would do. The easiest way of && together three return codes stored in variables is with the full $var -eq 0 && .... Surely there is a nicer way?

我知道bash并不是一种不错的编程语言-如果您甚至可以称呼它-但有什么办法可以使我不再那么尴尬?

I know bash is not a nice programming language - if you can even call it that - but is there any way I can make this less awkward?

推荐答案

您可以使用 bash 的算术命令对退出代码进行 OR 并取反，以获取退出代码1(如果其中任何一个代码非零).首先，一个例子:

You can use bash's arithmetic command to OR the exit codes together, and negate the result, to get an exit code of 1 if any of the codes is non-zero. First, an example:

$ ! (( 0 | 0 | 0 )); echo $?
0
$ ! (( 1 | 0 | 0 )); echo $?
1

现在，您的脚本:

#!/bin/bash

test1 ./src/ --test-1=option; exit_1=$?
test2 ./src/ test-2-options;  exit_2=$?   
test3 ./src/ -t 3 -o options; exit_3=$?

# Exit with error if any of the above failed. No need for a final
# call to exit, if this is the last command in the script
! (( $exit_1 || $exit_2 || $exit_3 ))

或者通常，您可以在运行任意数量的测试时累积退出代码:

Or in general, you can accumulate the exit codes as you run an arbitrary number of tests:

#!/bin/bash

# Unfortunately, ||= is not an assignment operator in bash.
# You could use |=, I suppose; you may not be able to assign
# any meaning to any particular non-zero value, though.
test1 ./src/ --test-1=option; (( exit_status = exit_status || $? ))
test2 ./src/ test-2-options;  (( exit_status = exit_status || $? ))  
test3 ./src/ -t 3 -o options; (( exit_status = exit_status || $? ))
# ...
testn ./src "${final_option_list[@]}"; (( exit_status = exit_status || $? ))

exit $exit_status   # 0 if they all succeeded, 1 if any failed

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