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    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:

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