进程调度 [英] Process Scheduling

问题描述

让我们说，我有我想作为cron作业定期运行10脚本。不过，我不希望他们都在同一时间运行。我想其中只有2同时运行。

Let's say, I have 10 scripts that I want to run regularly as cron jobs. However, I don't want all of them to run at the same time. I want only 2 of them running simultaneously.

一个解决方案，我想的是创建两个脚本，把5报表上的每个人，并把它们作为单独的crontab项。然而，解决方案看起来很即兴。

One solution that I'm thinking of is create two script, put 5 statements on each of them, and them as separate entries in the crontab. However the solution seem very adhoc.

有没有执行我上面提到的任务，现有的Unix工具？

Is there existing unix tool to perform the task I mentioned above?

推荐答案

的工作内置可以告诉你有多少孩子进程正在运行。一些简单的shell脚本可以完成这个任务：

The jobs builtin can tell you how many child processes are running. Some simple shell scripting can accomplish this task:

MAX_JOBS=2

launch_when_not_busy()
{
    while [ $(jobs | wc -l) -ge $MAX_JOBS ]
    do
        # at least $MAX_JOBS are still running.
        sleep 1
    done
    "$@" &
}

launch_when_not_busy  bash job1.sh --args
launch_when_not_busy  bash jobTwo.sh 
launch_when_not_busy  bash job_three.sh 
...
wait

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