Shell中实现“多线程”执行脚本文件完美解决方案
投稿：junjie
这篇文章主要介绍了Shell中实现“多线程”执行脚本文件完美解决方案,本文是针对一次调用执行多个脚本文件并分批执行的一个解决方法的测试和总结,需要的朋友可以参考下
即比如我有100个可执行文件，互相间没有特别的先后执行关系，如CODE:
复制代码 代码如下:
job_1
job_2
job_2
.....
job_100
想用csh/bash来多线程调用执行。
比如一次开5个线程，那么job_1,2,3,4,5一起先开始，那么其中任何一个线程如果先执行完成，则继续执行下一个没有初执行过的文件，如job_6,7,8....，这样一直以所指定的线程数来执行所有100个文件。
我本来想用 "&" 来放入后台，可是这样我一次可以指定5放入后台，但是无法知道其中任何一个程序何时执行完毕，所以也无法继续执行下一个程序啊!
完美解决方案：
复制代码 代码如下:
-(dearvoid@LinuxEden:Forum)-(~/tmp)-
[$$=6718 $?=0] ; cat job_1
#!/bin/bash
n=$((RANDOM % 5 + 1))
echo "$0 sleeping for $n seconds ..."
sleep $n
echo "$0 exiting ..."
-(dearvoid@LinuxEden:Forum)-(~/tmp)-
[$$=6718 $?=0] ; for ((i = 2; i <= 10; ++i)); do cp job_1 job_$i; done
-(dearvoid@LinuxEden:Forum)-(~/tmp)-
[$$=6718 $?=0] ; cat jobs.sh
#!/bin/bash
nParellel=5
nJobs=10
sJobPattern='./job_%d'
aJobs=()
sNextJob=
for ((iNextJob = 1; iNextJob <= nJobs; )); do
for ((iJob = 0; iJob < nParellel; ++iJob)); do
if [ $iNextJob -gt $nJobs ]; then
break;
fi
if [ ! "${aJobs[iJob]}" ] || ! kill -0 ${aJobs[iJob]} 2> /dev/null; then
printf -v sNextJob "$sJobPattern" $((iNextJob++))
echo "$sNextJob starting ..."
$sNextJob &
aJobs[iJob]=$!
fi
done
sleep .1
done
wait
-(dearvoid@LinuxEden:Forum)-(~/tmp)-
[$$=6718 $?=0] ; ./jobs.sh
./job_1 starting ...
./job_1 sleeping for 3 seconds ...
./job_2 starting ...
./job_2 sleeping for 2 seconds ...
./job_3 starting ...
./job_3 sleeping for 5 seconds ...
./job_4 starting ...
./job_5 starting ...
./job_4 sleeping for 4 seconds ...
./job_5 sleeping for 2 seconds ...
./job_2 exiting ...
./job_6 starting ...
./job_6 sleeping for 2 seconds ...
./job_5 exiting ...
./job_7 starting ...
./job_7 sleeping for 1 seconds ...
./job_1 exiting ...
./job_8 starting ...
./job_8 sleeping for 3 seconds ...
./job_7 exiting ...
./job_9 starting ...
./job_9 sleeping for 5 seconds ...
./job_4 exiting ...
./job_6 exiting ...
./job_10 starting ...
./job_10 sleeping for 5 seconds ...
./job_3 exiting ...
./job_8 exiting ...
./job_9 exiting ...
./job_10 exiting ...
-(dearvoid@LinuxEden:Forum)-(~/tmp)-
[$$=6718 $?=0] ; bye