从python运行程序,并在脚本被杀死后继续运行

2023-07-22Python开发问题
19

本文介绍了从python运行程序,并在脚本被杀死后继续运行的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

问题描述

我试过这样运行:

subprocess.Popen(['nohup', 'my_command'],
                 stdout=open('/dev/null', 'w'),
                 stderr=open('logfile.log', 'a'))

如果父脚本正常退出,这会起作用,但如果我终止脚本 (Ctrl-C),我的所有子进程也会被终止.有没有办法避免这种情况?

This works if the parent script exits gracefully, but if I kill the script (Ctrl-C), all my child processes are killed too. Is there a way to avoid this?

我关心的平台是 OS X 和 Linux,使用 Python 2.6 Python 2.7.

The platforms I care about are OS X and Linux, using Python 2.6 and Python 2.7.

推荐答案

在 Unix 系统上执行此操作的通常方法是,如果您是父级,则分叉并退出.看看 os.fork() .

The usual way to do this on Unix systems is to fork and exit if you're the parent. Have a look at os.fork() .

这是一个完成这项工作的函数:

Here's a function that does the job:

def spawnDaemon(func):
    # do the UNIX double-fork magic, see Stevens' "Advanced 
    # Programming in the UNIX Environment" for details (ISBN 0201563177)
    try: 
        pid = os.fork() 
        if pid > 0:
            # parent process, return and keep running
            return
    except OSError, e:
        print >>sys.stderr, "fork #1 failed: %d (%s)" % (e.errno, e.strerror) 
        sys.exit(1)

    os.setsid()

    # do second fork
    try: 
        pid = os.fork() 
        if pid > 0:
            # exit from second parent
            sys.exit(0) 
    except OSError, e: 
        print >>sys.stderr, "fork #2 failed: %d (%s)" % (e.errno, e.strerror) 
        sys.exit(1)

    # do stuff
    func()

    # all done
    os._exit(os.EX_OK)

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