- c - 在位数组中找到第一个零
- linux - Unix 显示有关匹配两种模式之一的文件的信息
- 正则表达式替换多个文件
- linux - 隐藏来自 xtrace 的命令
我正在运行一个 python 脚本,它可能需要也可能不需要几个小时才能完成。
在我的 python 脚本的开头,我想检查这个 python 脚本是否已经在运行。
如果它已经在运行,我想退出我刚刚启动的当前 python。
例如:
python 从凌晨 1 点开始,一直运行到凌晨 3 点在凌晨 2 点开始另一个,但不知道它已经在运行。我希望凌晨 2 点的 python 检查并退出,因为它已经在运行。
我该如何编写这个 python?
这是我尝试锁定的方法..
try:
l = lock.lock("/home/auto.py", timeout=600) # wait at most 10 minutes
except error.LockHeld:
e = sys.exc_info()[0]
logging.error("Error: " + str(e) + " at main gatering Stats")
smtpObj.sendmail(sender, receivers, message + "Error: " + str(e) + " at main gatering stats")
exit("Fail: " + str(e) + " at main gathering Stats")
else:
l.release()
所以我认为如果它仍在运行,将等待 10 分钟,然后退出。如果它不再运行,则运行当前的 python
最佳答案
您可以尝试使用 lockfile-create带有 r
标志的命令重试指定次数捕获 CalledProcessError
并退出,-p
标志将存储 pid
过程:
import os
import sys
from time import sleep
from subprocess import check_call, CalledProcessError
try:
check_call(["lockfile-create", "-q","-p", "-r", "0", "-l", "my.lock"])
except CalledProcessError as e:
print("{} is already running".format(sys.argv[0]))
print(e.returncode)
exit(1)
# main body
for i in range(10):
sleep(2)
print(1)
check_call(["rm","-f","my.lock"])
在已经运行的情况下使用上面的代码运行 test.py
脚本会输出以下内容:
$ python lock.py
lock.py is already running
4
选项
-q, --quiet
Suppress any output. Success or failure will only be indicated by the exit status.
-v, --verbose
Enable diagnostic output.
-l, --lock-name
Do not append .lock to the filename. This option applies to lockfile-create, lockfile-remove, lockfile-touch, or lockfile-check.
-p, --use-pid
Write the current process id (PID) to the lockfile whenever a lockfile is created, and use that pid when checking a lock's validity. See the lockfile_create(3) manpage for more information. This option applies to lockfile-create, lockfile-remove, lockfile-touch, and lockfile-check.
-o, --oneshot
Touch the lock and exit immediately. This option applies to lockfile-touch and mail-touchlock. When not provided, these commands will run forever, touching the lock once every minute until killed.
-r 重试次数, --retry 重试次数
Try to lock filename retry-count times before giving up. Each attempt will be delayed a bit longer than the last (in 5 second increments) until reaching a maximum delay of one minute between retries. If retry-count is unspecified, the default is 9 which will give up after 180 seconds (3 minutes) if all 9 lock attempts fail.
描述
The lockfile_create function creates a lockfile in an NFS safe way.
If flags is set to L_PID then lockfile_create will not only check for an existing lockfile, but it will read the contents as well to see if it contains a process id in ASCII. If so, the lockfile is only valid if that process still exists.
If the lockfile is on a shared filesystem, it might have been created by a process on a remote host. Thus the process-id checking is useless and the L_PID flag should not be set. In this case, there is no good way to see if a lockfile is stale. Therefore if the lockfile is older then 5 minutes, it will be removed. That is why the lockfile_touch function is provided: while holding the lock, it needs to be refreshed regularly (every minute or so) by calling lockfile_touch ().
The lockfile_check function checks if a valid lockfile is already present without trying to create a new lockfile.
Finally the lockfile_remove function removes the lockfile.
即使在 NFS 上,用于以原子方式创建锁定文件的算法如下:
1
A unique file is created. In printf format, the name of the file is .lk%05d%x%s. The first argument (%05d) is the current process id. The second argument (%x) consists of the 4 minor bits of the value returned by time(2). The last argument is the system hostname.
2
Then the lockfile is created using link(2). The return value of link is ignored.
3
Now the lockfile is stat()ed. If the stat fails, we go to step 6.
4
The stat value of the lockfile is compared with that of the temporary file. If they are the same, we have the lock. The temporary file is deleted and a value of 0 (success) is returned to the caller.
5
A check is made to see if the existing lockfile is a valid one. If it isn't valid, the stale lockfile is deleted.
6
Before retrying, we sleep for n seconds. n is initially 5 seconds, but after every retry 5 extra seconds is added up to a maximum of 60 seconds (an incremental backoff). Then we go to step 2 up to retries times.
似乎有一个等效的包叫做 lockfile-progs在 redhat 上。
在 Mac 上你可以使用 lockfile并做类似的事情:
import os
import sys
from time import sleep
import os
from subprocess import Popen, CalledProcessError, check_call
p = Popen(["lockfile", "-r", "0", "my.lock"])
p.wait()
if p.returncode == 0:
with open("my.pid", "w") as f:
f.write(str(os.getpid()))
else:
try:
with open("my.pid") as f:
# see if process is still running or lockfile
# is left over from previous run.
r = f.read()
check_call(["kill", "-0", "{}".format(r)])
except CalledProcessError:
# remove old lock file and create new
check_call(["rm", "-f", "my.lock"])
check_call(["lockfile", "-r", "0", "my.lock"])
# update pid
with open("my.pid", "w") as out:
out.write(str(os.getpid()))
print("Deleted stale lockfile.")
else:
print("{} is already running".format(sys.argv[0]))
print(p.returncode)
exit(1)
# main body
for i in range(10):
sleep(1)
print(1)
check_call(["rm", "-f", "my.lock"])
在您的情况下,也许可以使用套接字:
from socket import socket, gethostname, error, SO_REUSEADDR, SOL_SOCKET
from sys import argv
import errno
sock = socket()
# Create a socket object
host = gethostname()
# /proc/sys/net/ipv4/ip_local_port_range is 32768 61000 on my Ubuntu Machine
port = 60001
# allow connection in TIME_WAIT
sock.setsockopt(SOL_SOCKET, SO_REUSEADDR, 1)
try:
sock.bind((host, port))
sock.connect((host, port))
except error as e:
# [Errno 99] Cannot assign requested address
if e.errno == errno.EADDRNOTAVAIL:
print("{} is already running".format(argv[0]))
exit(1)
# else raise the error
else:
raise e
# main body
from time import sleep
while True:
print(1)
sleep(2)
sock.close()
关于python - 在 python 脚本中检查正在运行的 python 脚本,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29520587/
我正在处理一组标记为 160 个组的 173k 点。我想通过合并最接近的(到 9 或 10 个组)来减少组/集群的数量。我搜索过 sklearn 或类似的库,但没有成功。 我猜它只是通过 knn 聚类
我有一个扁平数字列表,这些数字逻辑上以 3 为一组,其中每个三元组是 (number, __ignored, flag[0 or 1]),例如: [7,56,1, 8,0,0, 2,0,0, 6,1,
我正在使用 pipenv 来管理我的包。我想编写一个 python 脚本来调用另一个使用不同虚拟环境(VE)的 python 脚本。 如何运行使用 VE1 的 python 脚本 1 并调用另一个 p
假设我有一个文件 script.py 位于 path = "foo/bar/script.py"。我正在寻找一种在 Python 中通过函数 execute_script() 从我的主要 Python
这听起来像是谜语或笑话,但实际上我还没有找到这个问题的答案。 问题到底是什么? 我想运行 2 个脚本。在第一个脚本中,我调用另一个脚本,但我希望它们继续并行,而不是在两个单独的线程中。主要是我不希望第
我有一个带有 python 2.5.5 的软件。我想发送一个命令,该命令将在 python 2.7.5 中启动一个脚本,然后继续执行该脚本。 我试过用 #!python2.7.5 和http://re
我在 python 命令行(使用 python 2.7)中,并尝试运行 Python 脚本。我的操作系统是 Windows 7。我已将我的目录设置为包含我所有脚本的文件夹,使用: os.chdir("
剧透:部分解决(见最后)。 以下是使用 Python 嵌入的代码示例: #include int main(int argc, char** argv) { Py_SetPythonHome
假设我有以下列表,对应于及时的股票价格: prices = [1, 3, 7, 10, 9, 8, 5, 3, 6, 8, 12, 9, 6, 10, 13, 8, 4, 11] 我想确定以下总体上最
所以我试图在选择某个单选按钮时更改此框架的背景。 我的框架位于一个类中,并且单选按钮的功能位于该类之外。 (这样我就可以在所有其他框架上调用它们。) 问题是每当我选择单选按钮时都会出现以下错误: co
我正在尝试将字符串与 python 中的正则表达式进行比较,如下所示, #!/usr/bin/env python3 import re str1 = "Expecting property name
考虑以下原型(prototype) Boost.Python 模块,该模块从单独的 C++ 头文件中引入类“D”。 /* file: a/b.cpp */ BOOST_PYTHON_MODULE(c)
如何编写一个程序来“识别函数调用的行号?” python 检查模块提供了定位行号的选项,但是, def di(): return inspect.currentframe().f_back.f_l
我已经使用 macports 安装了 Python 2.7,并且由于我的 $PATH 变量,这就是我输入 $ python 时得到的变量。然而,virtualenv 默认使用 Python 2.6,除
我只想问如何加快 python 上的 re.search 速度。 我有一个很长的字符串行,长度为 176861(即带有一些符号的字母数字字符),我使用此函数测试了该行以进行研究: def getExe
list1= [u'%app%%General%%Council%', u'%people%', u'%people%%Regional%%Council%%Mandate%', u'%ppp%%Ge
这个问题在这里已经有了答案: Is it Pythonic to use list comprehensions for just side effects? (7 个答案) 关闭 4 个月前。 告
我想用 Python 将两个列表组合成一个列表,方法如下: a = [1,1,1,2,2,2,3,3,3,3] b= ["Sun", "is", "bright", "June","and" ,"Ju
我正在运行带有最新 Boost 发行版 (1.55.0) 的 Mac OS X 10.8.4 (Darwin 12.4.0)。我正在按照说明 here构建包含在我的发行版中的教程 Boost-Pyth
学习 Python,我正在尝试制作一个没有任何第 3 方库的网络抓取工具,这样过程对我来说并没有简化,而且我知道我在做什么。我浏览了一些在线资源,但所有这些都让我对某些事情感到困惑。 html 看起来
我是一名优秀的程序员,十分优秀!