我编写了一个运行命令的函数,它以两个args为第一个命令,以秒为单位的第二个超时：
#! /bin/bash
function run_cmd {
cmd="$1"; timeout="$2"
grep -qP "^\d+$" <<< "$timeout" || timeout=10
stderrfile=$(readlink /proc/$$/fd/2)
exec 2<&-
exitfile=/tmp/exit_$(date +%s.%N)
(eval "$cmd";echo $? > $exitfile) &
start=$(date +%s)
while true; do
pid=$(jobs -l | awk '/Running/{print $2}')
if [ -n "$pid" ]; then
now=$(date +%s)
running=$(($now - $start))
if [ "$running" -ge "$timeout" ];then
kill -15 "$pid"
exit=1
fi
sleep 1
else
break
fi
done
test -n "$exit" || exit=$(cat $exitfile)
rm $exitfile
exec 2>$stderrfile
return "$exit"
}
function sleep5 {
sleep 5
echo "I slept 5"
return 2
}
run_cmd sleep5 "6"
run_cmd sleep5 "3"
echo "hi" >&2
该功能工作正常但我不确定它是一个优雅的解决方案,我想知道以下的替代品
>我必须将退出状态存储在文件中:( eval“$cmd”; echo $？> $exitfile)
>我正在关闭并重新打开STDERR：exec 2<& - 和exec 2> $stderrfile
我正在关闭STDERR因为我在杀死命令时无法避免该消息：
test.sh：第3行：32323终止(eval“$cmd”; echo $？> $exitfile)
PS：我知道超时和期望,但他们不适用于功能.
也许这适合您的需求.我更改了呼叫签名,以避免使用eval.
原文链接：https://www.f2er.com/bash/387182.html# Usage: run_with_timeout N cmd args...
# or: run_with_timeout cmd args...
# In the second case,cmd cannot be a number and the timeout will be 10 seconds.
run_with_timeout () {
local time=10
if [[ $1 =~ ^[0-9]+$]]; then time=$1; shift; fi
# Run in a subshell to avoid job control messages
( "$@" &
child=$!
# Avoid default notification in non-interactive shell for SIGTERM
trap -- "" SIGTERM
( sleep $time
kill $child 2> /dev/null ) &
wait $child
)
}
$sleep_and_exit() { sleep ${1:-1}; exit ${2:-0}; }
$time run_with_timeout 1 sleep_and_exit 3 0; echo $?
real 0m1.007s
user 0m0.003s
sys 0m0.006s
143
$time run_with_timeout 3 sleep_and_exit 1 0; echo $?
real 0m1.007s
user 0m0.003s
sys 0m0.008s
0
$time run_with_timeout 3 sleep_and_exit 1 7; echo $?
real 0m1.006s
user 0m0.001s
sys 0m0.006s
7
如图所示,run_with_timeout的退出状态将是执行命令的退出状态,除非它被超时杀死,在这种情况下它将是143(128 15).
注意：如果设置了一个大的超时和/或运行了一个forkbomb,你可能会以足够快的速度回收pid,kill-child会杀死错误的进程.
