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?
当前回答
如果你用的是Unix，
import signal
...
class Alarm(Exception):
pass
def alarm_handler(signum, frame):
raise Alarm
signal.signal(signal.SIGALRM, alarm_handler)
signal.alarm(5*60) # 5 minutes
try:
stdoutdata, stderrdata = proc.communicate()
signal.alarm(0) # reset the alarm
except Alarm:
print "Oops, taking too long!"
# whatever else
其他回答
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()
我已经修改了sussudio的答案。现在函数返回:(returncode, stdout, stderr, timeout) - stdout和stderr被解码为utf-8字符串
def kill_proc(proc, timeout):
timeout["value"] = True
proc.kill()
def run(cmd, timeout_sec):
proc = subprocess.Popen(shlex.split(cmd), stdout=subprocess.PIPE, stderr=subprocess.PIPE)
timeout = {"value": False}
timer = Timer(timeout_sec, kill_proc, [proc, timeout])
timer.start()
stdout, stderr = proc.communicate()
timer.cancel()
return proc.returncode, stdout.decode("utf-8"), stderr.decode("utf-8"), timeout["value"]
如果你用的是Unix，
import signal
...
class Alarm(Exception):
pass
def alarm_handler(signum, frame):
raise Alarm
signal.signal(signal.SIGALRM, alarm_handler)
signal.alarm(5*60) # 5 minutes
try:
stdoutdata, stderrdata = proc.communicate()
signal.alarm(0) # reset the alarm
except Alarm:
print "Oops, taking too long!"
# whatever else
没想到居然没人提到超时
Timeout 5 ping -c 3 somehost
显然，这并不适用于每个用例，但如果您处理的是一个简单的脚本，那么这是很难克服的。
mac用户也可以通过homebrew在coreutils中使用gtimeout。
对于python 2.6+，使用gevent
from gevent.subprocess import Popen, PIPE, STDOUT
def call_sys(cmd, timeout):
p= Popen(cmd, shell=True, stdout=PIPE)
output, _ = p.communicate(timeout=timeout)
assert p.returncode == 0, p. returncode
return output
call_sys('./t.sh', 2)
# t.sh example
sleep 5
echo done
exit 1
