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?
当前回答
陷阱是你的朋友。在很多系统中都可能出现ERR。您可以捕获EXIT，或在DEBUG上在每个命令之后执行一段代码。
这除了所有的标准信号。
edit
这是一次意外的登录错误的帐户，所以我没有看到示例的请求。
试试这里，用我的普通账户。
在bash脚本中处理异常
其他回答
set -e
fail () {
touch .failure
}
expect () {
wait
if [ -f .failure ]; then
rm -f .failure
exit 1
fi
}
sleep 2 || fail &
sleep 2 && false || fail &
sleep 2 || fail
expect
顶部的set -e使脚本在失败时停止。
如果任何子作业失败，Expect将返回1。
捕获CHLD信号可能不起作用，因为如果它们同时到达，您可能会丢失一些信号。
#!/bin/bash
trap 'rm -f $tmpfile' EXIT
tmpfile=$(mktemp)
doCalculations() {
echo start job $i...
sleep $((RANDOM % 5))
echo ...end job $i
exit $((RANDOM % 10))
}
number_of_jobs=10
for i in $( seq 1 $number_of_jobs )
do
( trap "echo job$i : exit value : \$? >> $tmpfile" EXIT; doCalculations ) &
done
wait
i=0
while read res; do
echo "$res"
let i++
done < "$tmpfile"
echo $i jobs done !!!
这是有效的，应该是一个很好的，如果不是更好的@HoverHell的答案!
#!/usr/bin/env bash
set -m # allow for job control
EXIT_CODE=0; # exit code of overall script
function foo() {
echo "CHLD exit code is $1"
echo "CHLD pid is $2"
echo $(jobs -l)
for job in `jobs -p`; do
echo "PID => ${job}"
wait ${job} || echo "At least one test failed with exit code => $?" ; EXIT_CODE=1
done
}
trap 'foo $? $$' CHLD
DIRN=$(dirname "$0");
commands=(
"{ echo "foo" && exit 4; }"
"{ echo "bar" && exit 3; }"
"{ echo "baz" && exit 5; }"
)
clen=`expr "${#commands[@]}" - 1` # get length of commands - 1
for i in `seq 0 "$clen"`; do
(echo "${commands[$i]}" | bash) & # run the command via bash in subshell
echo "$i ith command has been issued as a background job"
done
# wait for all to finish
wait;
echo "EXIT_CODE => $EXIT_CODE"
exit "$EXIT_CODE"
# end
当然，我已经在一个NPM项目中保存了这个脚本，它允许你并行运行bash命令，对测试很有用:
https://github.com/ORESoftware/generic-subshell
从Bash 5.1开始，由于引入了wait -p，有了一种很好的等待和处理多个后台作业结果的新方法:
#!/usr/bin/env bash
# Spawn background jobs
for ((i=0; i < 10; i++)); do
secs=$((RANDOM % 10)); code=$((RANDOM % 256))
(sleep ${secs}; exit ${code}) &
echo "Started background job (pid: $!, sleep: ${secs}, code: ${code})"
done
# Wait for background jobs, print individual results, determine overall result
result=0
while true; do
wait -n -p pid; code=$?
[[ -z "${pid}" ]] && break
echo "Background job ${pid} finished with code ${code}"
(( ${code} != 0 )) && result=1
done
# Return overall result
exit ${result}
这是我目前想到的。我想了解如何在子进程终止时中断sleep命令，这样就不必根据使用情况调优WAITALL_DELAY。
waitall() { # PID...
## Wait for children to exit and indicate whether all exited with 0 status.
local errors=0
while :; do
debug "Processes remaining: $*"
for pid in "$@"; do
shift
if kill -0 "$pid" 2>/dev/null; then
debug "$pid is still alive."
set -- "$@" "$pid"
elif wait "$pid"; then
debug "$pid exited with zero exit status."
else
debug "$pid exited with non-zero exit status."
((++errors))
fi
done
(("$#" > 0)) || break
# TODO: how to interrupt this sleep when a child terminates?
sleep ${WAITALL_DELAY:-1}
done
((errors == 0))
}
debug() { echo "DEBUG: $*" >&2; }
pids=""
for t in 3 5 4; do
sleep "$t" &
pids="$pids $!"
done
waitall $pids
