Shell中实现“多线程”执行脚本文件完美解决方案
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:
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：
-(dearvoid@LinuxEden:Forum)-(~/tmp)-
[$$=6718$?=0];catjob_1
#!/bin/bash
n=$((RANDOM%5+1))
echo"$0sleepingfor$nseconds..."
sleep$n
echo"$0exiting..."
-(dearvoid@LinuxEden:Forum)-(~/tmp)-
[$$=6718$?=0];for((i=2;i<=10;++i));docpjob_1job_$i;done
-(dearvoid@LinuxEden:Forum)-(~/tmp)-
[$$=6718$?=0];catjobs.sh
#!/bin/bash
nParellel=5
nJobs=10
sJobPattern='./job_%d'
aJobs=()
sNextJob=
for((iNextJob=1;iNextJob<=nJobs;));do
for((iJob=0;iJob<nParellel;++iJob));do
if[$iNextJob-gt$nJobs];then
break;
fi
if[!"${aJobs[iJob]}"]||!kill-0${aJobs[iJob]}2>/dev/null;then
printf-vsNextJob"$sJobPattern"$((iNextJob++))
echo"$sNextJobstarting..."
$sNextJob&
aJobs[iJob]=$!
fi
done
sleep.1
done
wait
-(dearvoid@LinuxEden:Forum)-(~/tmp)-
[$$=6718$?=0];./jobs.sh
./job_1starting...
./job_1sleepingfor3seconds...
./job_2starting...
./job_2sleepingfor2seconds...
./job_3starting...
./job_3sleepingfor5seconds...
./job_4starting...
./job_5starting...
./job_4sleepingfor4seconds...
./job_5sleepingfor2seconds...
./job_2exiting...
./job_6starting...
./job_6sleepingfor2seconds...
./job_5exiting...
./job_7starting...
./job_7sleepingfor1seconds...
./job_1exiting...
./job_8starting...
./job_8sleepingfor3seconds...
./job_7exiting...
./job_9starting...
./job_9sleepingfor5seconds...
./job_4exiting...
./job_6exiting...
./job_10starting...
./job_10sleepingfor5seconds...
./job_3exiting...
./job_8exiting...
./job_9exiting...
./job_10exiting...
-(dearvoid@LinuxEden:Forum)-(~/tmp)-
[$$=6718$?=0];bye