Shell“多线程”执行脚本文件 Shell中实现“多线程”执行脚本文件完美解决方案
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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
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