Python在一定時間內運行子進程

Question

我正在測試某個exe文件，並且想為我的腳本實現一種確定它已進入無限循環的方法。 這是我當前的代碼：

import subprocess
import os
import sys
runs = 1000 # Default run is 1000
if len(sys.argv)>1: # If I want to change the num of runs
    runs = int(sys.argv[1])
FNULL = open(os.devnull, 'w')
logfile = open('logfile', 'w')
args = "exe" # Exe to test
succ = 0
fail = 0
for i in range (0,runs):
    if subprocess.call(args,stdout = logfile, stderr = FNULL) == 100:
        succ += 1 # If returned 100 then success
    else:
        fail += 1 # Else Failed
        break # Break on failure
    open('logfile', 'w').close() # Empties the file
print "Succ: %d , Fail: %d" % (succ, fail)

可以說我定義了一個無限循環，因為我的exe運行時間超過5秒。 我將如何實施呢？ 感謝您提供任何幫助，包括當前代碼的提示！

Answer 1

啟動一個threading.Timer ，它將在5秒鍾后終止該進程，並報告該行為已完成。 您需要在不同的步驟中創建並等待該過程，因此請使用Popen對象而不是call 。 我創建了一個測試程序，該程序使用sleep模擬您的無限列表。

import subprocess
import os
import sys
import threading

def on_timeout(proc, status_dict):
    """Kill process on timeout and note as status_dict['timeout']=True"""
    # a container used to pass status back to calling thread
    status_dict['timeout'] = True
    print("timed out")
    proc.kill()

runs = 1000 # Default run is 1000
if len(sys.argv)>1: # If I want to change the num of runs
    runs = int(sys.argv[1])
FNULL = open(os.devnull, 'w')
logfile = open('logfile', 'w')
# replacing example with a running program. This is a simple python
# we can call from the command line.
# args = "exe" # Exe to test
test_script = "import time;import sys;time.sleep(%d);sys.exit(100)"
succ = 0
fail = 0
for i in range (0,runs):
    # set by timer 
    status_dict = {'timeout':False}
    # test prog sleeps i seconds
    args = ["python", "-c", test_script % i]
    proc = subprocess.Popen(args, 
        stdout = logfile, stderr = FNULL)
    # trigger timout and kill process in 5 seconds
    timer = threading.Timer(5, on_timeout, (proc, status_dict))
    timer.start()
    proc.wait()
    # in case we didn't hit timeout
    timer.cancel()
    print status_dict
    if not status_dict['timeout'] and proc.returncode == 100:
        succ += 1 # If returned 100 then success
    else:
        fail += 1 # Else Failed
        break # Break on failure
    open('logfile', 'w').close() # Empties the file
print "Succ: %d , Fail: %d" % (succ, fail)

Answer 2

這對我有用

import subprocess
import time

for _ in range(999999): # or whatever
    p1 = subprocess.Popen(['exe_file', 'arg1', 'arg2'])
    time.sleep(0.1)  # wait a small amount of time, hopefully the process ends fast
    return_code = p1.poll()

    if return_code is not None: # we're happy, the process ended fast
        ... # do something to celebrate
    else:  # we're sad. let the process run for at most 5 seconds
        time.sleep(5)   
        p1.terminate()    # this kills the process. try p1.kill() too...
        p1.wait()   #  this cleans up the process registry.

免責聲明：這是linux代碼。 Windows上的情況可能有所不同，但是您可以先檢查一下，然后在這里https://docs.python.org/2/library/subprocess.html閱讀有關子進程庫以及linux和Windows之間差異的內容。

Answer 3

在python3.3中，超時被添加到subprocess.call中。 如果您使用的是python3.3，則可以更改subprocess.call以將超時作為參數：

   subprocess.call(args,stdout = logfile, stderr = FNULL, timeout=5)

如果您使用的是python2.7，則可以使用subprocess32軟件包，或者需要編寫一些額外的代碼來處理超時。

如果安裝subprocess32模塊，則可以使用上述subprocess.call方法，並將超時作為參數。

另外，此代碼可以幫助您實現相同的功能：

from subprocess import Popen
timeout=5 #5 seconds
p = Popen(args, shell = True, stdout = logfile, stderr = FNULL)
while (p.poll() is None and timeout > 0):
    time.sleep(1)
    timeout-=1

if timeout <= 0:
    p.terminate()  #Timeout