在bash中添加(收集)退出代码

问题描述

我需要依赖于脚本中的几个单独的执行，并且不想将它们全部捆绑在一个丑陋的"if"语句中.我想使用退出代码"$?"每个执行并添加它；最后，如果该值超过阈值-我想执行一条命令.

I need to depend on few separate executions in a script and don't want to bundle them all in an ugly 'if' statement. I would like to take the exit code '$?' of each execution and add it; at the end, if this value is over a threshold - I would like to execute a command.

伪代码:

ALLOWEDERROR=5

run_something
RESULT=$?
..other things..

run_something_else
RESULT=$RESULT + $?

if [ $RESULT -gt ALLOWEDERROR ]
   then echo "Too many errors"
fi

问题:即使互联网上有其他要求，bash也拒绝对待RESULT和$?作为整数.正确的语法是什么?

Issue: Even though the Internet claims otherwise, bash refuses to treat the RESULT and $? as integer. What is the correct syntax?

谢谢.

推荐答案

您可能想看一下内置的 trap 看看是否有帮助:

You might want to take a look at the trap builtin to see if it would be helpful:

help trap

或

man bash

您可以为此类错误设置陷阱:

you can set a trap for errors like this:

#!/bin/bash

AllowedError=5

SomeErrorHandler () {
    (( errcount++ ))       # or (( errcount += $? ))
    if  (( errcount > $AllowedError ))
    then
        echo "Too many errors"
        exit $errcount
    fi
}

trap SomeErrorHandler ERR

for i in {1..6}
do
    false
    echo "Reached $i"     # "Reached 6" is never printed
done

echo "completed"          # this is never printed

如果您像这样计算错误(并且仅当它们是错误时)，而不是使用" $?"，那么您不必担心返回值不是零或一.例如，单个返回值127会使您立即超过阈值.除了 ERR ，您还可以为其他信号注册 trap .

If you count the errors (and only when they are errors) like this instead of using "$?", then you don't have to worry about return values that are other than zero or one. A single return value of 127, for example, would throw you over your threshold immediately. You can also register traps for other signals in addition to ERR.

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