Python只允许同一个py文件执行一个process

在计划任务使用周期性的任务的时候，经常会遇到一个问题py文件同时启动了很多次，造成windows系统中进程很多，并且查看进程名称都是python.exe。以下代码仅限windows

测试在WinXP成功，Other OS 没有测试过。

import os

import re

import sys

 

 

def getProcessInfo(processName):

    command ="tasklist | findstr \""+processName[0]+"\""

    ret = os.system(command)

    if ret != 0:        #can't find it

        #print processName[0]+" can't find it"

        return False

    else:                #find it

        #print processName[0]+" find it"

        command = 'wmic process where caption="'+processName[0]+'" get caption,commandline /value'

        pipe = os.popen(command)

        pipeString = pipe.read()

        pipe.close()

        #matchObj = re.compile(r"CommandLine=.+?"+processName[0]+" (.+?)\r\n",re.I)

        matchObj = re.compile(r"CommandLine=.+?python.exe['\"]{1}[\\t ]*[\"']{1}(.+?)[\"']{1}",re.I)

        #CommandLine=.+?python.exe['"]{1}[\t ]*["']{1}(.+?)["']{1}

        list = matchObj.findall(pipeString)

        number = 0

        print pipeString

        for i in list:

            print i

            if i == processName[1]:

                #print processName[0]+" "+processName[1]+" find it"

                number=number+1

                

        if number >= 2:

            return True

        else:

            #print processName[1]+" can't find it"

            return False

 

def getProcessCount(processName):

    command ='tasklist | findstr "'+processName[0]+'"'

 

#print getProcessInfo(["python.exe","-k imgsvc"])         #-k imgsvc

if getProcessInfo(["python.exe",sys.argv[0]]) == True:      #is process myself

    print sys.argv[0]+" is alway running!!!"

    os.system("pause")

else:

    print sys.argv[0]+" isn't running!!!"

    os.system("pause")

如果在程序中调用最后几行的函式，就能够很好的解决前面提到的这个问题；发现自己是否还正在运行中。。