python子进程一个一个运行

Question

i have some sort of processes:

subprocess.Popen(['python2.7 script1.py')],shell=True)
subprocess.Popen(['python2.7 script2.py')],shell=True)
subprocess.Popen(['python2.7 script3.py')],shell=True)
subprocess.Popen(['python2.7 script4.py')],shell=True)

i want to each one starts after the previous process completely finish. i mean

subprocess.Popen(['python2.7 script2.py')],shell=True)

starts after

subprocess.Popen(['python2.7 script1.py')],shell=True)

completly finished, and for others the same. this is cause previous scripts has output that it is used by next script. thanks

Answer 1

You can simply use wait() for each one to finish, like this:

sp1 = subprocess.Popen(['python2.7 script1.py'],shell=True)
sp1.wait()

sp2 = subprocess.Popen(['python2.7 script2.py'],shell=True)
sp2.wait()

sp3 = subprocess.Popen(['python2.7 script3.py'],shell=True)
sp3.wait()

sp4 = subprocess.Popen(['python2.7 script4.py'],shell=True)
sp4.wait()

Or in shorter way:

subprocess.Popen(['python2.7 script1.py'],shell=True).wait()
subprocess.Popen(['python2.7 script2.py'],shell=True).wait()
subprocess.Popen(['python2.7 script3.py'],shell=True).wait()
subprocess.Popen(['python2.7 script4.py'],shell=True).wait()

Answer 2

Use subprocess.call :

Run the command described by args . Wait for command to complete, then return the returncode attribute.

In your example:

subprocess.call(['python2.7 script1.py'],shell=True)
subprocess.call(['python2.7 script2.py'],shell=True)
subprocess.call(['python2.7 script3.py'],shell=True)
subprocess.call(['python2.7 script4.py'],shell=True)