问题
I tried these two methods:
os.system("python test.py")
subprocess.Popen("python test.py", shell=True)
Both approaches need to wait until test.py finishes which blocks main process. I know "nohup" can do the job. Is there a Python way to launch test.py or any other shell scripts and leave it running in background?
Suppose test.py is like this:
for i in range(0, 1000000):
print i
Both os.system() or subprocess.Popen() will block main program until 1000000 lines of output displayed. What I want is let test.py runs silently and display main program output only. Main program may quie while test.py is still running.
回答1:
subprocess.Popen(["python", "test.py"]) should work.
Note that the job might still die when your main script exits. In this case, try subprocess.Popen(["nohup", "python", "test.py"])
回答2:
os.spawnlp(os.P_NOWAIT, "path_to_test.py", "test.py")
来源：https://stackoverflow.com/questions/1605520/how-to-launch-and-run-external-script-in-background