Python popen命令.等到命令完成

Question

我有一个脚本,我用popen启动一个shell命令.问题是脚本不会等到popen命令完成后立即继续.

om_points = os.popen(command, "w")
.....

如何判断我的Python脚本要等到shell命令完成？

Answer 1

根据您希望如何使用脚本,您有两种选择.如果您希望命令在执行时阻止而不执行任何操作,则可以使用subprocess.call.

#start and block until done
subprocess.call([data["om_points"], ">", diz['d']+"/points.xml"])

如果您想在执行或进行操作时执行操作stdin,可以communicate在popen通话后使用.

#start and process things, then wait
p = subprocess.Popen([data["om_points"], ">", diz['d']+"/points.xml"])
print "Happens while running"
p.communicate() #now wait plus that you can send commands to process

如文档中所述,wait可能会死锁,因此建议进行通信.

Answer 2

你可以用它subprocess来实现这一目标.

import subprocess

#This command could have multiple commands separated by a new line \n
some_command = "export PATH=$PATH://server.sample.mo/app/bin \n customupload abc.txt"

p = subprocess.Popen(some_command, stdout=subprocess.PIPE, shell=True)

(output, err) = p.communicate()  

#This makes the wait possible
p_status = p.wait()

#This will give you the output of the command being executed
print "Command output: " + output

Answer 3

popen通过执行以下操作强制在读取所有输出之前不继续：

os.popen(command).read()

Answer 4

让您尝试传递的命令成为

os.system('x')

然后你把它转换成一个声明

t = os.system('x')

现在，python 将等待命令行的输出，以便将其分配给变量t。

Answer 5

wait()对我来说效果很好。子进程p1、p2和p3同时执行。因此，所有过程在3秒后完成。

import subprocess

processes = []

p1 = subprocess.Popen("sleep 3", stdout=subprocess.PIPE, shell=True)
p2 = subprocess.Popen("sleep 3", stdout=subprocess.PIPE, shell=True)
p3 = subprocess.Popen("sleep 3", stdout=subprocess.PIPE, shell=True)

processes.append(p1)
processes.append(p2)
processes.append(p3)

for p in processes:
    if p.wait() != 0:
        print("There was an error")

print("all processed finished")

Answer 6

您正在寻找的是wait方法.