我想抑制运行可执行文件的函数产生的所有终端输出.
我试图通过使用每次调用函数时临时重新定义stdout和stderr的上下文管理器来抑制Python函数的输出.这会抑制函数中的print调用产生的终端输出,但是当函数调用产生终端输出的可执行文件时,它似乎不起作用.
那么,如何抑制Python函数调用的可执行文件的输出？
我的代码如下.我已经包含了一个示例函数,它调用ls来试图说明我想要抑制的终端输出的类型(虽然我正在处理的函数是不同的).
#!/usr/bin/env python
import os
import subprocess
import sys
def main():
print("hello")
with silence():
print("there")
print("world")
with silence():
engage_command(command = "ls")
class silence(object):
def __init__(
self,
stdout = None,
stderr = None
):
if stdout == None and stderr == None:
devnull = open(os.devnull, "w")
stdout = devnull
stderr = devnull
self._stdout = stdout or sys.stdout
self._stderr = stderr or sys.stderr
def __enter__(
self
):
self.old_stdout = sys.stdout
self.old_stderr = sys.stderr
self.old_stdout.flush()
self.old_stderr.flush()
sys.stdout = self._stdout
sys.stderr = self._stderr
def __exit__(
self,
exc_type,
exc_value,
traceback
):
self._stdout.flush()
self._stderr.flush()
sys.stdout = self.old_stdout
sys.stderr = self.old_stderr
def engage_command(
command = None
):
process = subprocess.Popen(
[command],
shell = True,
executable = "/bin/bash")
process.wait()
output, errors = process.communicate()
return output
if __name__ == "__main__":
main()
在我的特定情况下,我正在尝试运行以下函数(而不是上面的ls函数)：
with propyte.silence():
stream = pyaudio.PyAudio().open(
format = pyaudio.PyAudio().get_format_from_width(1),
channels = 1,
rate = bitrate,
output = True
)
运行时,会产生如下输出：
ALSA lib pcm_dsnoop.c:606:(snd_pcm_dsnoop_open) unable to open slave
ALSA lib pcm_dmix.c:1029:(snd_pcm_dmix_open) unable to open slave
ALSA lib pcm.c:2266:(snd_pcm_open_noupdate) Unknown PCM cards.pcm.rear
ALSA lib pcm.c:2266:(snd_pcm_open_noupdate) Unknown PCM cards.pcm.center_lfe
ALSA lib pcm.c:2266:(snd_pcm_open_noupdate) Unknown PCM cards.pcm.side
ALSA lib pcm_dmix.c:1029:(snd_pcm_dmix_open) unable to open slave
Cannot connect to server socket err = No such file or directory
Cannot connect to server request channel
jack server is not running or cannot be started
JackShmReadWritePtr::~JackShmReadWritePtr - Init not done for 4294967295, skipping unlock
JackShmReadWritePtr::~JackShmReadWritePtr - Init not done for 4294967295, skipping unlock
我想压制那个输出.
编辑：测试@Matthias提供的a solution
#!/usr/bin/env python
import contextlib
import os
import subprocess
import sys
def main():
print("hello")
with silence():
print("there")
print("world")
with silence():
engage_command(command = "ls")
@contextlib.contextmanager
def silence():
devnull = os.open(os.devnull, os.O_WRONLY)
old_stderr = os.dup(2)
sys.stderr.flush()
os.dup2(devnull, 2)
os.close(devnull)
try:
yield
finally:
os.dup2(old_stderr, 2)
os.close(old_stderr)
def engage_command(
command = None
):
process = subprocess.Popen(
[command],
shell = True,
executable = "/bin/bash")
process.wait()
output, errors = process.communicate()
return output
if __name__ == "__main__":
main()
我没有成功地抑制打印或ls的终端输出,我不知道为什么.
