为什么 shell=True 会吃掉我的 subprocess.Popen 标准输出? [英] Why does shell=True eat my subprocess.Popen stdout?

问题描述

似乎在链的第一个进程中使用 shell=True 会以某种方式从下游任务中删除标准输出:

It seems that using shell=True in the first process of a chain somehow drops the stdout from downstream tasks:

p1 = Popen(['echo','hello'], stdout=PIPE)
p2 = Popen('cat', stdin=p1.stdout, stdout=PIPE)
p2.communicate()
# outputs correctly ('hello
', None)

让第一个进程使用 shell=True 会以某种方式杀死输出...

Making the first process use shell=True kills the output somehow...

p1 = Popen(['echo','hello'], stdout=PIPE, shell=True)
p2 = Popen('cat', stdin=p1.stdout, stdout=PIPE)
p2.communicate()
# outputs incorrectly ('
', None)

shell=True 在第二个进程上似乎并不重要.这是预期的行为吗?

shell=True on the second process doesn't seem to matter. Is this expected behavior?

推荐答案

当你传递 shell=True 时，Popen 需要一个字符串参数，而不是一个列表.所以当你这样做时:

When you pass shell=True, Popen expects a single string argument, not a list. So when you do this:

p1 = Popen(['echo','hello'], stdout=PIPE, shell=True)

会发生什么:

execve("/bin/sh", ["/bin/sh", "-c", "echo", "hello"], ...)

也就是说，它调用 sh -c echo"，而 hello 被有效地忽略(从技术上讲，它成为 shell 的位置参数).所以 shell 运行 echo，它打印 ，这就是为什么你会在输出中看到它.

That is, it calls sh -c "echo", and hello is effectively ignored (technically it becomes a positional argument to the shell). So the shell runs echo, which prints , which is why you see that in your output.

如果你使用shell=True，你需要这样做:

If you use shell=True, you need to do this:

p1 = Popen('echo hello', stdout=PIPE, shell=True)

这篇关于为什么 shell=True 会吃掉我的 subprocess.Popen 标准输出?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！