蟒子由

Question

运行一个我有某种的过程：

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) 

我想每个人开始后前面的过程完全结束。 我的意思是

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

开始

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

完全地结束后，和其他人一样。这是因为以前的脚本已经输出了它被下一个脚本使用的输出。 感谢

Answer 1

您可以简单地使用wait()为每一个来完成，像这样：

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() 

或者更短的方式：

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

使用subprocess.call：

运行命令描述参数。等待命令完成，然后返回returncode属性。

在您的例子：

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)