子进程在 PyCharm 中正确显示 sys.stdout 但不在控制台中 [英] Subprocess displays sys.stdout correctly in PyCharm but not in the console

问题描述

我有一个小的 Python 程序，它从一个名为 commandwrapper(它是 subprocess.popen 的包装器)的 pip 包中执行一个终端命令:https://pypi.python.org/pypi/commandwrapper/0.7).我还试图将实时输出捕获到控制台和文件中.

I have a small Python program that executes a terminal command from a pip package called commandwrapper (which is a wrapper for subprocess.popen: https://pypi.python.org/pypi/commandwrapper/0.7). I am also trying to capture the real-time output to the console and to a file.

我有代码:

class Tee(object):
    def __init__(self, *files):
        self.files = files
    def write(self, obj):
        for f in self.files:
            f.write(obj)
            f.flush()
    def flush(self) :
        for f in self.files:
            f.flush()

# Set the stdout/stderr to Tee()
out = open(stdout_log, 'w')
stdout_reset = sys.stdout
sys.stdout = Tee(sys.stdout, out)

process = commandwrapper.WrapCommand(command, shell=True) # Also tried False with same behaivor.
cmd = process.prepareToRun()

# Stream the output to the console and stdoutfiles
while cmd.poll() is None:
    msg_out = cmd.stdout.readline()
    sys.stdout.write(msg_out)
    sys.stdout.flush()

out.close()

当我在 PyCharm 中运行它时，这非常有效.command 的输出被写入文件并实时显示在终端控制台上.

This works perfect when I run it in PyCharm. The output of command is written to the file AND displayed on the terminal console in real-time.

但是，当我在终端中运行相同的代码时，控制台上没有显示任何输出.怎么会这样?stdout 在文件中被正确捕获，但没有写入控制台.

However, when I run the same code in a terminal, no output is displayed on the console. How can this be? The stdout is correctly captured in a file, but nothing is written to the console.

谁能看出这段代码在 PyCharm 中运行良好并符合预期的任何原因，但没有向终端显示任何标准输出?我在这里不知所措.如果有的话，如果行为被逆转，我可以处理它.

Can anyone see any reason that this code would work well and as expected in PyCharm, but not display any stdout to the terminal? I'm at a loss here. If anything, I could deal with it if the behavior was reversed.

使用 OSX Yosemite 并运行 bash.

Using OSX Yosemite and running bash.

推荐答案

您需要更改投票的逻辑，我使用了 Popen，但如果您愿意，也可以使用包装器:

You need to change the logic where you have poll, I used Popen but you can use the wrapper if you prefer:

out = open(stdout_log, 'w')
stdout_reset = sys.stdout
sys.stdout = Tee(sys.stdout, out)
from subprocess import Popen,PIPE,STDOUT
process = Popen([list_of_cmds],stdout=PIPE,stderr=STDOUT)
# Stream the output to the console and stdoutfiles
for line in iter(process.stdout.readline,""):
    sys.stdout.write(line)


out.close()

应用相同的逻辑适用于 commandwrapper 库:

Applying the same logic works with the commandwrapper lib:

process = commandwrapper.WrapCommand(command, shell=True) # Also tried False with same behaivor.
cmd = process.prepareToRun()
# Stream the output to the console and stdoutfiles
for line in iter(cmd.stdout.readline,""):
    sys.stdout.write(line)

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