在没有挂起的情况下停止在 Python 中读取进程输出?
问题描述
我有一个适用于 Linux 的 Python 程序，几乎看起来像这样:
I have a Python program for Linux almost looks like this one :
import os
import time
process = os.popen("top").readlines()
time.sleep(1)
os.popen("killall top")
print process
程序挂在这一行:
process = os.popen("top").readlines()
这发生在不断更新输出的工具中，例如Top"
and that happens in the tools that keep update outputting like "Top"
我最好的尝试:
import os
import time
import subprocess
process = subprocess.Popen('top')
time.sleep(2)
os.popen("killall top")
print process
它比第一个效果更好(它被 kelled 了)，但它返回了:
it worked better than the first one (it's kelled ), but it returns :
<subprocess.Popen object at 0x97a50cc>
二审:
import os
import time
import subprocess
process = subprocess.Popen('top').readlines()
time.sleep(2)
os.popen("killall top")
print process
与第一个相同.由于readlines()"而挂起
the same as the first one. It hanged due to "readlines()"
它的返回应该是这样的:
Its returning should be like this :
top - 05:31:15 up 12:12, 5 users, load average: 0.25, 0.14, 0.11
Tasks: 174 total, 2 running, 172 sleeping, 0 stopped, 0 zombie
Cpu(s): 9.3%us, 3.8%sy, 0.1%ni, 85.9%id, 0.9%wa, 0.0%hi, 0.0%si, 0.0%st
Mem: 1992828k total, 1849456k used, 143372k free, 233048k buffers
Swap: 4602876k total, 0k used, 4602876k free, 1122780k cached
PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND
31735 Barakat 20 0 246m 52m 20m S 19.4 2.7 13:54.91 totem
1907 root 20 0 91264 45m 15m S 1.9 2.3 38:54.14 Xorg
2138 Barakat 20 0 17356 5368 4284 S 1.9 0.3 3:00.15 at-spi-registry
2164 Barakat 9 -11 164m 7372 6252 S 1.9 0.4 2:54.58 pulseaudio
2394 Barakat 20 0 27212 9792 8256 S 1.9 0.5 6:01.48 multiload-apple
6498 Barakat 20 0 56364 30m 18m S 1.9 1.6 0:03.38 pyshell
1 root 20 0 2880 1416 1208 S 0.0 0.1 0:02.02 init
2 root 20 0 0 0 0 S 0.0 0.0 0:00.02 kthreadd
3 root RT 0 0 0 0 S 0.0 0.0 0:00.12 migration/0
4 root 20 0 0 0 0 S 0.0 0.0 0:02.07 ksoftirqd/0
5 root RT 0 0 0 0 S 0.0 0.0 0:00.00 watchdog/0
9 root 20 0 0 0 0 S 0.0 0.0 0:01.43 events/0
11 root 20 0 0 0 0 S 0.0 0.0 0:00.00 cpuset
12 root 20 0 0 0 0 S 0.0 0.0 0:00.02 khelper
13 root 20 0 0 0 0 S 0.0 0.0 0:00.00 netns
14 root 20 0 0 0 0 S 0.0 0.0 0:00.00 async/mgr
15 root 20 0 0 0 0 S 0.0 0.0 0:00.00 pm
并保存在变量进程"中.任何我的想法伙计们，我现在真的被困住了吗?
and save in the variable "process". Any I idea guys, I'm really stuck now ?
解决方案
#!/usr/bin/env python
"""Start process; wait 2 seconds; kill the process; print all process output."""
import subprocess
import tempfile
import time
def main():
# open temporary file (it automatically deleted when it is closed)
# `Popen` requires `f.fileno()` so `SpooledTemporaryFile` adds nothing here
f = tempfile.TemporaryFile()
# start process, redirect stdout
p = subprocess.Popen(["top"], stdout=f)
# wait 2 seconds
time.sleep(2)
# kill process
#NOTE: if it doesn't kill the process then `p.wait()` blocks forever
p.terminate()
p.wait() # wait for the process to terminate otherwise the output is garbled
# print saved output
f.seek(0) # rewind to the beginning of the file
print f.read(),
f.close()
if __name__=="__main__":
main()
只打印输出部分的类似尾巴的解决方案
您可以在另一个线程中读取进程输出并将所需数量的最后一行保存在队列中:
Tail-like Solutions that print only the portion of the output
You could read the process output in another thread and save the required number of the last lines in a queue:
import collections
import subprocess
import time
import threading
def read_output(process, append):
for line in iter(process.stdout.readline, ""):
append(line)
def main():
# start process, redirect stdout
process = subprocess.Popen(["top"], stdout=subprocess.PIPE, close_fds=True)
try:
# save last `number_of_lines` lines of the process output
number_of_lines = 200
q = collections.deque(maxlen=number_of_lines) # atomic .append()
t = threading.Thread(target=read_output, args=(process, q.append))
t.daemon = True
t.start()
#
time.sleep(2)
finally:
process.terminate() #NOTE: it doesn't ensure the process termination
# print saved lines
print ''.join(q)
if __name__=="__main__":
main()
这个变体要求 q.append() 是原子操作.否则输出可能会损坏.
This variant requires q.append() to be atomic operation. Otherwise the output might be corrupted.
您可以使用 signal.alarm() 在指定超时后调用 process.terminate() 而不是在另一个线程中读取.尽管它可能无法与 subprocess 模块很好地交互.基于 @Alex Martelli 的回答:
You could use signal.alarm() to call the process.terminate() after specified timeout instead of reading in another thread. Though it might not interact very well with the subprocess module. Based on @Alex Martelli's answer:
import collections
import signal
import subprocess
class Alarm(Exception):
pass
def alarm_handler(signum, frame):
raise Alarm
def main():
# start process, redirect stdout
process = subprocess.Popen(["top"], stdout=subprocess.PIPE, close_fds=True)
# set signal handler
signal.signal(signal.SIGALRM, alarm_handler)
signal.alarm(2) # produce SIGALRM in 2 seconds
try:
# save last `number_of_lines` lines of the process output
number_of_lines = 200
q = collections.deque(maxlen=number_of_lines)
for line in iter(process.stdout.readline, ""):
q.append(line)
signal.alarm(0) # cancel alarm
except Alarm:
process.terminate()
finally:
# print saved lines
print ''.join(q)
if __name__=="__main__":
main()
这种方法只适用于 *nix 系统.如果 process.stdout.readline() 没有返回，它可能会阻塞.
This approach works only on *nix systems. It might block if process.stdout.readline() doesn't return.
import collections
import subprocess
import threading
def main():
# start process, redirect stdout
process = subprocess.Popen(["top"], stdout=subprocess.PIPE, close_fds=True)
# terminate process in timeout seconds
timeout = 2 # seconds
timer = threading.Timer(timeout, process.terminate)
timer.start()
# save last `number_of_lines` lines of the process output
number_of_lines = 200
q = collections.deque(process.stdout, maxlen=number_of_lines)
timer.cancel()
# print saved lines
print ''.join(q),
if __name__=="__main__":
main()
这种方法也应该适用于 Windows.在这里，我使用 process.stdout 作为可迭代对象；它可能会引入额外的输出缓冲，如果不需要，您可以切换到 iter(process.stdout.readline, "") 方法.如果进程没有在 process.terminate() 上终止，则脚本挂起.
This approach should also work on Windows. Here I've used process.stdout as an iterable; it might introduce an additional output buffering, you could switch to the iter(process.stdout.readline, "") approach if it is not desirable. if the process doesn't terminate on process.terminate() then the scripts hangs.
import collections
import subprocess
import sys
import time
def main():
args = sys.argv[1:]
if not args:
args = ['top']
# start process, redirect stdout
process = subprocess.Popen(args, stdout=subprocess.PIPE, close_fds=True)
# save last `number_of_lines` lines of the process output
number_of_lines = 200
q = collections.deque(maxlen=number_of_lines)
timeout = 2 # seconds
now = start = time.time()
while (now - start) < timeout:
line = process.stdout.readline()
if not line:
break
q.append(line)
now = time.time()
else: # on timeout
process.terminate()
# print saved lines
print ''.join(q),
if __name__=="__main__":
main()
此变体既不使用线程，也不使用信号，但会在终端中产生乱码输出.如果 process.stdout.readline() 阻塞，它将阻塞.
This variant use neither threads, no signals but it produces garbled output in the terminal. It will block if process.stdout.readline() blocks.
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