蟒蛇： os.spawn 。无法在后台启动 bash 进程
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python: os.spawn . cannot start bash process on background
提问by Daniel Gurianov
Task is to execute bash script from python script and let it execute on background, even if python script will finish. I need UNIX solution and i do not care if it will be not working on Win.
任务是从 python 脚本执行 bash 脚本并让它在后台执行，即使 python 脚本将完成。我需要 UNIX 解决方案，我不在乎它是否不适用于 Win。
Python script :
Python脚本：
#!/usr/bin/env python
import os, commands
command = '/usr/bin/ssh localhost "/home/gd/test/python/back.sh " '
print os.spawnlp(os.P_NOWAIT,command)
print "Python done"
/home/gd/test/python/back.sh :
/home/gd/test/python/back.sh ：
#!/usr/bin/bash
/bin/echo "started"
/bin/sleep 80
/bin/echo "ended"
The issue is, when python script starts , i see PID of spawned process printed. But there is no process on background. When i use P_WAIT i see exit code 127 which means that command not found in the path. But i already provided all paths that already possible? These scripts works perfectly with commands.getouput.
问题是，当 python 脚本启动时，我看到打印的生成进程的 PID。但是后台没有进程。当我使用 P_WAIT 时，我看到退出代码 127，这意味着在路径中找不到该命令。但是我已经提供了所有可能的路径？这些脚本与 commands.getouput 完美配合。
回答by John La Rooy
Something like this should work
这样的事情应该工作
#!/usr/bin/env python
import os
command = ['/usr/bin/ssh', 'ssh', 'localhost', '/home/gd/test/python/back.sh']
print os.spawnlp(os.P_NOWAIT, *command)
print "Python done"
Note that it's preferable to use the subprocessmodule here instead of spawn
请注意，这里最好使用subprocess模块而不是 spawn
#!/usr/bin/env python
from subprocess import Popen
command = ['/usr/bin/ssh', 'localhost', '/home/gd/test/python/back.sh']
print Popen(command)
print "Python done"
