如何在bash脚本中等待从该脚本派生的几个子进程完成,然后在任何子进程以code !=0结束时返回退出代码!=0?
简单的脚本:
#!/bin/bash
for i in `seq 0 9`; do
doCalculations $i &
done
wait
上面的脚本将等待所有10个子进程,但它总是给出退出状态0(参见help wait)。我如何修改这个脚本,以便它将发现衍生子进程的退出状态,并在任何子进程以code !=0结束时返回退出代码1 ?
有没有比收集子进程的pid、按顺序等待它们并求和退出状态更好的解决方案呢?
我最近用过这个(感谢Alnitak):
#!/bin/bash
# activate child monitoring
set -o monitor
# locking subprocess
(while true; do sleep 0.001; done) &
pid=$!
# count, and kill when all done
c=0
function kill_on_count() {
# you could kill on whatever criterion you wish for
# I just counted to simulate bash's wait with no args
[ $c -eq 9 ] && kill $pid
c=$((c+1))
echo -n '.' # async feedback (but you don't know which one)
}
trap "kill_on_count" CHLD
function save_status() {
local i=$1;
local rc=$2;
# do whatever, and here you know which one stopped
# but remember, you're called from a subshell
# so vars have their values at fork time
}
# care must be taken not to spawn more than one child per loop
# e.g don't use `seq 0 9` here!
for i in {0..9}; do
(doCalculations $i; save_status $i $?) &
done
# wait for locking subprocess to be killed
wait $pid
echo
从这里,我们可以很容易地推断,并拥有一个触发器(触摸文件,发送信号)并改变计数标准(计数触摸的文件,或其他)以响应该触发器。或者如果你只是想要'any'非零rc,只需从save_status中杀死锁。