下面是运行任意命令返回其标准输出数据的Python代码,或在非零退出码上引发异常:
proc = subprocess.Popen(
cmd,
stderr=subprocess.STDOUT, # Merge stdout and stderr
stdout=subprocess.PIPE,
shell=True)
communication用于等待进程退出:
stdoutdata, stderrdata = proc.communicate()
子进程模块不支持超时——杀死运行超过X秒的进程的能力——因此,通信可能需要很长时间才能运行。
在Windows和Linux上运行的Python程序中实现超时的最简单方法是什么?
timeout现在由子进程模块中的call()和communication()支持(从Python3.3开始):
import subprocess
subprocess.call("command", timeout=20, shell=True)
这将调用该命令并引发异常
subprocess.TimeoutExpired
如果命令在20秒后还没有完成。
然后你可以处理异常来继续你的代码,就像这样:
try:
subprocess.call("command", timeout=20, shell=True)
except subprocess.TimeoutExpired:
# insert code here
希望这能有所帮助。
另一种选择是写入临时文件以防止标准输出阻塞,而不需要使用communication()轮询。在其他答案没有的地方,这个方法对我有用;比如在窗户上。
outFile = tempfile.SpooledTemporaryFile()
errFile = tempfile.SpooledTemporaryFile()
proc = subprocess.Popen(args, stderr=errFile, stdout=outFile, universal_newlines=False)
wait_remaining_sec = timeout
while proc.poll() is None and wait_remaining_sec > 0:
time.sleep(1)
wait_remaining_sec -= 1
if wait_remaining_sec <= 0:
killProc(proc.pid)
raise ProcessIncompleteError(proc, timeout)
# read temp streams from start
outFile.seek(0);
errFile.seek(0);
out = outFile.read()
err = errFile.read()
outFile.close()
errFile.close()
我有一个问题,我想终止一个多线程子进程,如果它花费的时间超过给定的超时长度。我想在Popen()中设置一个超时,但它不起作用。然后,我意识到Popen().wait()等于call(),所以我有了在.wait(timeout=xxx)方法中设置超时的想法,这最终工作了。因此,我是这样解决的:
import os
import sys
import signal
import subprocess
from multiprocessing import Pool
cores_for_parallelization = 4
timeout_time = 15 # seconds
def main():
jobs = [...YOUR_JOB_LIST...]
with Pool(cores_for_parallelization) as p:
p.map(run_parallel_jobs, jobs)
def run_parallel_jobs(args):
# Define the arguments including the paths
initial_terminal_command = 'C:\\Python34\\python.exe' # Python executable
function_to_start = 'C:\\temp\\xyz.py' # The multithreading script
final_list = [initial_terminal_command, function_to_start]
final_list.extend(args)
# Start the subprocess and determine the process PID
subp = subprocess.Popen(final_list) # starts the process
pid = subp.pid
# Wait until the return code returns from the function by considering the timeout.
# If not, terminate the process.
try:
returncode = subp.wait(timeout=timeout_time) # should be zero if accomplished
except subprocess.TimeoutExpired:
# Distinguish between Linux and Windows and terminate the process if
# the timeout has been expired
if sys.platform == 'linux2':
os.kill(pid, signal.SIGTERM)
elif sys.platform == 'win32':
subp.terminate()
if __name__ == '__main__':
main()
This solution kills the process tree in case of shell=True, passes parameters to the process (or not), has a timeout and gets the stdout, stderr and process output of the call back (it uses psutil for the kill_proc_tree). This was based on several solutions posted in SO including jcollado's. Posting in response to comments by Anson and jradice in jcollado's answer. Tested in Windows Srvr 2012 and Ubuntu 14.04. Please note that for Ubuntu you need to change the parent.children(...) call to parent.get_children(...).
def kill_proc_tree(pid, including_parent=True):
parent = psutil.Process(pid)
children = parent.children(recursive=True)
for child in children:
child.kill()
psutil.wait_procs(children, timeout=5)
if including_parent:
parent.kill()
parent.wait(5)
def run_with_timeout(cmd, current_dir, cmd_parms, timeout):
def target():
process = subprocess.Popen(cmd, cwd=current_dir, shell=True, stdout=subprocess.PIPE, stdin=subprocess.PIPE, stderr=subprocess.PIPE)
# wait for the process to terminate
if (cmd_parms == ""):
out, err = process.communicate()
else:
out, err = process.communicate(cmd_parms)
errcode = process.returncode
thread = Thread(target=target)
thread.start()
thread.join(timeout)
if thread.is_alive():
me = os.getpid()
kill_proc_tree(me, including_parent=False)
thread.join()