使用os.kill（）后子流程何时终止？ [英] How to find out when subprocess has terminated after using os.kill()?
问题描述
subprocess.Popen 。由于我的应用程序的架构限制，我无法使用 Popen.terminate（） 终止子进程， Popen.poll（） 来检查进程何时终止。这是因为我不能在变量中引用启动的子进程。 相反，我必须编写进程标识 pid 到子程序启动时的文件 pidfile 。当我想停止子进程时，我打开这个 pidfile 并使用 os.kill（pid，signal.SIGTERM） 来阻止它。
我的问题是：如何查看子流程何时终止？在调用 os.kill（）之后，使用 signal.SIGTERM 需要大约1-2分钟才能终止。首先，我认为 os.waitpid（） 将是这个任务的正确的事情，但是当我在 os.kill（）之后调用它时，它给我 OSError：[Errno 10]没有子进程。
顺便说一下，我使用两种窗体启动和停止来自HTML模板的子进程，程序逻辑在Django视图中。当我的应用程序处于调试模式时，该异常会在浏览器中显示。知道我在视图中调用的子进程（ python manage.py crawlwebpages ）本身也可以调用另一个子进程，即Scrapy抓取器的一个实例。我将这个Scrapy实例的 pid 写入 pidfile ，这是我要终止的。
以下是相关代码：
def process_main_page_forms（request）：
如果request.method =='POST'：
if request.POST ['form-type'] == u'webpage-crawler-form'：
template_context = _crawl_webpage（request）
elif request.POST ['form-type'] == u'stop-crawler-form'：
template_context = _stop_crawler（request）
else：
template_context = {
'webpage_crawler_form'：WebPageCrawlerForm（），
'stop_crawler_form'：StopCrawlerForm（）}
返回渲染（request，'main.html'，template_context）
def _crawl_webpage（request）：
webpage_crawler_form = WebPageCrawlerForm（request.POST）
如果webpage_crawler_form.is_valid（）：
url_to_crawl = webpage_crawler _form.cleaned_data ['url_to_crawl']
maximum_pages_to_crawl = webpage_crawler_form.cleaned_data ['maximum_pages_to_crawl']
program ='python manage.py crawlwebpages'+'-n'+ str（maximum_pages_to_crawl）+ ''+ url_to_crawl
p = subprocess.Popen（program.split（））
template_context = {
'webpage_crawler_form'：webpage_crawler_form，
'stop_crawler_form'：StopCrawlerForm ）
返回template_context
def _stop_crawler（请求）：
stop_crawler_form = StopCrawlerForm（request.POST）
如果stop_crawler_form.is_valid （）：
with open（'scrapy_crawler_process.pid'，'rb'）as pidfile：
process_id = int（pidfile.read（）。strip（））
print'PROCESS ID： '，process_id
os.kill（process_id，signal.SIGTERM）
os.waitpid（process_id，os.WNOHANG）＃这给我的OSError
print'Crawler process terminated！'
template_context = {
'webpage_crawler_form'：WebPageCrawlerForm（），
'stop_crawler_form'：stop_crawler_form}
return template_context
我该怎么办？非常感谢！
编辑：
根据由 Jacek Konieczny ，我可以通过将函数 _stop_crawler（request）中的代码更改为以下内容来解决我的问题：
def _stop_crawler（request）：
stop_crawler_form = StopCrawlerForm（request.POST）
如果stop_crawler_form.is_valid（）：
打开（'scrapy_crawler_process.pid'，'rb'）作为pidfile：
process_id = int（pidfile.read（）。strip（））
＃这些是基本行
os.kill（process_id，signal.SIGTERM）
while True：
try：
time.sleep（10）
os.kill（process_id，0）
除OSError：
break
print'Crawler procedures ss已终止！'
template_context = {
'webpage_crawler_form'：WebPageCrawlerForm（），
'stop_crawler_form'：stop_crawler_form}
返回template_context
检查进程是否仍在运行的通常方法是杀死（）它与信号'0'。对于正在运行的作业，它不会执行任何操作，并且如果进程不存在，则会引发 OSError 异常与 errno = ESRCH / p>
[jajcus @ lolek〜] $ sleep 1000&
[1] 2405
[jajcus @ lolek〜] $ python
Python 2.7.3（默认，2012年5月11日，11:57:22）
[GCC 4.6.3 20120315（release）] on linux2
有关详细信息，请输入help，copyright，credits或license。
>>> import os
>>> os.kill（2405，0）
>>>> os.kill（2405，15）
>>>> os.kill（2405，0）
追溯（最近的最后一次呼叫）：
文件< stdin>，第1行，< module>
OSError：[Errno 3]没有这样的过程
但是，只要有可能，调用进程的父进程，并使用 wait（）函数系列来处理其终止。那就是 Popen 对象。
I have a Python program (precisely, a Django application) that starts a subprocess using subprocess.Popen. Due to architecture constraints of my application, I'm not able to use Popen.terminate() to terminate the subprocess and Popen.poll() to check when the process has terminated. This is because I cannot hold a reference to the started subprocess in a variable.
Instead, I have to write the process id pid to a file pidfile when the subprocess starts. When I want to stop the subprocess, I open this pidfile and use os.kill(pid, signal.SIGTERM) to stop it.
My question is: How can I find out when the subprocess has really terminated? Using signal.SIGTERM it needs approximately 1-2 minutes to finally terminate after calling os.kill(). First I thought that os.waitpid() would be the right thing for this task but when I call it after os.kill() it gives me OSError: [Errno 10] No child processes.
By the way, I'm starting and stopping the subprocess from a HTML template using two forms and the program logic is inside a Django view. The exception gets displayed in my browser when my application is in debug mode. It's probably also important to know that the subprocess that I call in my view (python manage.py crawlwebpages) itself calls another subprocess, namely an instance of a Scrapy crawler. I write the pid of this Scrapy instance to the pidfile and this is what I want to terminate.
Here is the relevant code:
def process_main_page_forms(request):
if request.method == 'POST':
if request.POST['form-type'] == u'webpage-crawler-form':
template_context = _crawl_webpage(request)
elif request.POST['form-type'] == u'stop-crawler-form':
template_context = _stop_crawler(request)
else:
template_context = {
'webpage_crawler_form': WebPageCrawlerForm(),
'stop_crawler_form': StopCrawlerForm()}
return render(request, 'main.html', template_context)
def _crawl_webpage(request):
webpage_crawler_form = WebPageCrawlerForm(request.POST)
if webpage_crawler_form.is_valid():
url_to_crawl = webpage_crawler_form.cleaned_data['url_to_crawl']
maximum_pages_to_crawl = webpage_crawler_form.cleaned_data['maximum_pages_to_crawl']
program = 'python manage.py crawlwebpages' + ' -n ' + str(maximum_pages_to_crawl) + ' ' + url_to_crawl
p = subprocess.Popen(program.split())
template_context = {
'webpage_crawler_form': webpage_crawler_form,
'stop_crawler_form': StopCrawlerForm()}
return template_context
def _stop_crawler(request):
stop_crawler_form = StopCrawlerForm(request.POST)
if stop_crawler_form.is_valid():
with open('scrapy_crawler_process.pid', 'rb') as pidfile:
process_id = int(pidfile.read().strip())
print 'PROCESS ID:', process_id
os.kill(process_id, signal.SIGTERM)
os.waitpid(process_id, os.WNOHANG) # This gives me the OSError
print 'Crawler process terminated!'
template_context = {
'webpage_crawler_form': WebPageCrawlerForm(),
'stop_crawler_form': stop_crawler_form}
return template_context
What can I do? Thank you very much!
EDIT:
According to the great answer given by Jacek Konieczny, I could solve my problem by changing my code in the function _stop_crawler(request) to the following:
def _stop_crawler(request):
stop_crawler_form = StopCrawlerForm(request.POST)
if stop_crawler_form.is_valid():
with open('scrapy_crawler_process.pid', 'rb') as pidfile:
process_id = int(pidfile.read().strip())
# These are the essential lines
os.kill(process_id, signal.SIGTERM)
while True:
try:
time.sleep(10)
os.kill(process_id, 0)
except OSError:
break
print 'Crawler process terminated!'
template_context = {
'webpage_crawler_form': WebPageCrawlerForm(),
'stop_crawler_form': stop_crawler_form}
return template_context
The usual way to check if a process is still running is to kill() it with signal '0'. It does nothing to a running job and raises an OSError exception with errno=ESRCH if the process does not exist.
[jajcus@lolek ~]$ sleep 1000 &
[1] 2405
[jajcus@lolek ~]$ python
Python 2.7.3 (default, May 11 2012, 11:57:22)
[GCC 4.6.3 20120315 (release)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import os
>>> os.kill(2405, 0)
>>> os.kill(2405, 15)
>>> os.kill(2405, 0)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
OSError: [Errno 3] No such process
But whenever possible the caller should stay a parent of the called process and use wait() function family to handle its termination. That is what Popen object does.
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