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 $i &
done
wait
上面的脚本将等待所有10个子进程,但它总是给出退出状态0(参见help wait)。我如何修改这个脚本,以便它将发现衍生子进程的退出状态,并在任何子进程以code !=0结束时返回退出代码1 ?
有没有比收集子进程的pid、按顺序等待它们并求和退出状态更好的解决方案呢?
正是为了这个目的,我写了一个bash函数:for。
注意::for不仅保留并返回失败函数的退出码,而且终止所有并行运行的实例。在这种情况下可能不需要。
#!/usr/bin/env bash
# Wait for pids to terminate. If one pid exits with
# a non zero exit code, send the TERM signal to all
# processes and retain that exit code
#
# usage:
# :wait 123 32
function :wait(){
local pids=("$@")
[ ${#pids} -eq 0 ] && return $?
trap 'kill -INT "${pids[@]}" &>/dev/null || true; trap - INT' INT
trap 'kill -TERM "${pids[@]}" &>/dev/null || true; trap - RETURN TERM' RETURN TERM
for pid in "${pids[@]}"; do
wait "${pid}" || return $?
done
trap - INT RETURN TERM
}
# Run a function in parallel for each argument.
# Stop all instances if one exits with a non zero
# exit code
#
# usage:
# :for func 1 2 3
#
# env:
# FOR_PARALLEL: Max functions running in parallel
function :for(){
local f="${1}" && shift
local i=0
local pids=()
for arg in "$@"; do
( ${f} "${arg}" ) &
pids+=("$!")
if [ ! -z ${FOR_PARALLEL+x} ]; then
(( i=(i+1)%${FOR_PARALLEL} ))
if (( i==0 )) ;then
:wait "${pids[@]}" || return $?
pids=()
fi
fi
done && [ ${#pids} -eq 0 ] || :wait "${pids[@]}" || return $?
}
使用
for.sh:
#!/usr/bin/env bash
set -e
# import :for from gist: https://gist.github.com/Enteee/c8c11d46a95568be4d331ba58a702b62#file-for
# if you don't like curl imports, source the actual file here.
source <(curl -Ls https://gist.githubusercontent.com/Enteee/c8c11d46a95568be4d331ba58a702b62/raw/)
msg="You should see this three times"
:(){
i="${1}" && shift
echo "${msg}"
sleep 1
if [ "$i" == "1" ]; then sleep 1
elif [ "$i" == "2" ]; then false
elif [ "$i" == "3" ]; then
sleep 3
echo "You should never see this"
fi
} && :for : 1 2 3 || exit $?
echo "You should never see this"
$ ./for.sh; echo $?
You should see this three times
You should see this three times
You should see this three times
1
参考文献
[1]:博客
[2]:要点
从Bash 5.1开始,由于引入了wait -p,有了一种很好的等待和处理多个后台作业结果的新方法:
#!/usr/bin/env bash
# Spawn background jobs
for ((i=0; i < 10; i++)); do
secs=$((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}