Python: 每 x 分钟交替调用函数
假设我有下面这四个函数:
def foo():
subprocess.Popen('start /B someprogramA.exe', shell=True)
def bar():
subprocess.Popen('start /B someprogramB.exe', shell=True)
def foo_kill():
subprocess.Popen('taskkill /IM someprogramA.exe')
def bar_kill():
subprocess.Popen('taskkill /IM someprogramB.exe')
我该如何让这两个函数 foo 和 bar 每隔30分钟交替运行一次呢?
意思是:第一个30分钟运行 foo
,第二个30分钟运行 bar
,第三个30分钟再运行 foo
,以此类推。每次新的运行应该“结束”上一个线程或函数。
我有一个倒计时的线程,但不太确定怎么“交替”这些函数。
class Timer(threading.Thread):
def __init__(self, minutes):
self.runTime = minutes
threading.Thread.__init__(self)
class CountDownTimer(Timer):
def run(self):
counter = self.runTime
for sec in range(self.runTime):
#do something
time.sleep(60) #editted from 1800 to 60 - sleeps for a minute
counter -= 1
timeout=30
c=CountDownTimer(timeout)
c.start()
编辑:这是我根据 Nicholas Knight 的建议找到的解决方案……
import threading
import subprocess
import time
timeout=2 #alternate delay gap in minutes
def foo():
subprocess.Popen('start /B notepad.exe', shell=True)
def bar():
subprocess.Popen('start /B calc.exe', shell=True)
def foo_kill():
subprocess.Popen('taskkill /IM notepad.exe')
def bar_kill():
subprocess.Popen('taskkill /IM calc.exe')
class Alternator(threading.Thread):
def __init__(self, timeout):
self.delay_mins = timeout
self.functions = [(foo, foo_kill), (bar, bar_kill)]
threading.Thread.__init__(self)
def run(self):
while True:
for f, kf in self.functions:
f()
time.sleep(self.delay_mins*60)
kf()
a=Alternator(timeout)
a.start()
运行得很好。
4 个回答
2
用一个变量来记录你上次运行的是哪个函数。当计时器触发时,运行另一个函数,并更新这个变量。
10
记住,在Python中,函数是第一类对象。这意味着你可以把它们存储在变量和容器里!一种实现的方法是:
funcs = [(foo, foo_kill), (bar, bar_kill)]
def run(self):
counter = self.runTime
for sec in range(self.runTime):
runner, killer = funcs[counter % 2] # the index alternates between 0 and 1
runner() # do something
time.sleep(1800)
killer() # kill something
counter -= 1
6
你把这个搞得太复杂了。
while True:
foo()
time.sleep(1800)
foo_kill()
bar()
time.sleep(1800)
bar_kill()
或者如果你想以后更方便地添加更多功能:
functions = [(foo, foo_kill), (bar, bar_kill), ] # Just append more as needed
while True:
for f, kf in functions:
f()
time.sleep(1800)
kf()