Python 초간단 크론 중복 실행 방지

tendo 모듈 설치

easy_install tendo

pip install tendo

from tendo import singleton
me = singleton.SingleInstance()

기존 스크립트에 위 내용만 추가 해주면 되고, 인스턴스가 실행중이면 에러 발생 시키고 sys.exit(-1) 됨

최신 singleton.py 스트립트는 https://github.com/pycontribs/tendo/blob/master/tendo/singleton.py 여기서 확인하면 됨

#! /usr/bin/env python
 
import logging
from multiprocessing import Process
import os
import sys
import tempfile
import unittest
 
 
if sys.platform != "win32":
    import fcntl
 
 
class SingleInstanceException(BaseException):
    pass
 
 
class SingleInstance(object):
 
    """Class that can be instantiated only once per machine.
    If you want to prevent your script from running in parallel just instantiate SingleInstance() class. If is there another instance already running it will throw a `SingleInstanceException`.
    >>> import tendo
    ... me = SingleInstance()
    This option is very useful if you have scripts executed by crontab at small amounts of time.
    Remember that this works by creating a lock file with a filename based on the full path to the script file.
    Providing a flavor_id will augment the filename with the provided flavor_id, allowing you to create multiple singleton instances from the same file. This is particularly useful if you want specific functions to have their own singleton instances.
    """
 
    def __init__(self, flavor_id="", lockfile=""):
        self.initialized = False
        if lockfile:
            self.lockfile = lockfile
        else:
            basename = os.path.splitext(os.path.abspath(sys.argv[0]))[0].replace(
                "/", "-").replace(":", "").replace("\\", "-") + '-%s' % flavor_id + '.lock'
            self.lockfile = os.path.normpath(
                tempfile.gettempdir() + '/' + basename)
 
        logger.debug("SingleInstance lockfile: " + self.lockfile)
        if sys.platform == 'win32':
            try:
                # file already exists, we try to remove (in case previous
                # execution was interrupted)
                if os.path.exists(self.lockfile):
                    os.unlink(self.lockfile)
                self.fd = os.open(
                    self.lockfile, os.O_CREAT | os.O_EXCL | os.O_RDWR)
            except OSError:
                type, e, tb = sys.exc_info()
                if e.errno == 13:
                    logger.error(
                        "Another instance is already running, quitting.")
                    raise SingleInstanceException()
                print(e.errno)
                raise
        else:  # non Windows
            self.fp = open(self.lockfile, 'w')
            self.fp.flush()
            try:
                fcntl.lockf(self.fp, fcntl.LOCK_EX | fcntl.LOCK_NB)
            except IOError:
                logger.warning(
                    "Another instance is already running, quitting.")
                raise SingleInstanceException()
        self.initialized = True
 
    def __del__(self):
        if not self.initialized:
            return
        try:
            if sys.platform == 'win32':
                if hasattr(self, 'fd'):
                    os.close(self.fd)
                    os.unlink(self.lockfile)
            else:
                fcntl.lockf(self.fp, fcntl.LOCK_UN)
                # os.close(self.fp)
                if os.path.isfile(self.lockfile):
                    os.unlink(self.lockfile)
        except Exception as e:
            if logger:
                logger.warning(e)
            else:
                print("Unloggable error: %s" % e)
            sys.exit(-1)
 
 
def f(name):
    tmp = logger.level
    logger.setLevel(logging.CRITICAL)  # we do not want to see the warning
    try:
        me2 = SingleInstance(flavor_id=name)  # noqa
    except SingleInstanceException:
        sys.exit(-1)
    logger.setLevel(tmp)
    pass
 
 
class testSingleton(unittest.TestCase):
 
    def test_1(self):
        me = SingleInstance(flavor_id="test-1")
        del me  # now the lock should be removed
        assert True
 
    def test_2(self):
        p = Process(target=f, args=("test-2",))
        p.start()
        p.join()
        # the called function should succeed
        assert p.exitcode == 0, "%s != 0" % p.exitcode
 
    def test_3(self):
        me = SingleInstance(flavor_id="test-3")  # noqa -- me should still kept
        p = Process(target=f, args=("test-3",))
        p.start()
        p.join()
        # the called function should fail because we already have another
        # instance running
        assert p.exitcode != 0, "%s != 0 (2nd execution)" % p.exitcode
        # note, we return -1 but this translates to 255 meanwhile we'll
        # consider that anything different from 0 is good
        p = Process(target=f, args=("test-3",))
        p.start()
        p.join()
        # the called function should fail because we already have another
        # instance running
        assert p.exitcode != 0, "%s != 0 (3rd execution)" % p.exitcode
 
    def test_4(self):
        lockfile = '/tmp/foo.lock'
        me = SingleInstance(lockfile=lockfile)
        assert me.lockfile == lockfile
 
 
logger = logging.getLogger("tendo.singleton")
 
if __name__ == "__main__":
    logger.addHandler(logging.StreamHandler())
    logger.setLevel(logging.DEBUG)
    unittest.main()