Question

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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.

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Solution

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"])

OTHER TIPS

os.spawnlp(os.P_NOWAIT, "path_to_test.py", "test.py")
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