bash - bash函数非零退出代码

Question

在下面的函数“handleExit”中，exitCode为零，即“0”。我希望它是“1”。我的函数称为“退出1”。为了检测功能“writeToErr”具有非零状态，我需要做什么。

#!/bin/bash

set -euo pipefail

logError() {
  awk " BEGIN { print \"$@\" > \"/dev/fd/2\" }"
}

function writeToErr ()  {
    echo "standard out"
    logError "standard err"
    exit 1
}

function wrapper () {
    writeToErr >>output.log 2>&1
}

function handleExit () {
    echo "handle exit"
    exitCode=$?
    if [ $exitCode -eq "0" ]
     then
        echo "No problem"
     else
        echo "$0 exited unexpectedly with status:$exitCode"
        exit 1
     fi
}

trap wrapper EXIT
handleExit >>output.log 2>&1

这是“output.log”的内容:

handle exit
No problem
standard out
standard err

Answer 1

有两个问题:

您在运行handleExit之前先运行wrapper，所以它尚未失败。

handleExit检查echo的退出代码

没有描述您要做什么，我想您想要:

#!/bin/bash

set -euo pipefail

logError() {
    # A better way to write to stderr
    echo "$*" >&2
}

function writeToErr ()  {
    echo "standard out"
    logError "standard err"
    exit 1
}

function wrapper () {
    writeToErr >>output.log 2>&1
}

function handleExit () {
    # Get exit code of the previous command, instead of echo
    exitCode=$?
    echo "handle exit"
    if [ $exitCode -eq "0" ]
     then
        echo "No problem"
     else
        echo "$0 exited unexpectedly with status:$exitCode"
        exit 1
     fi
}

# Call handler on exit, so it has something to handle
trap handleExit EXIT
wrapper