如何让我的函数在python中每秒运行一次?

时间:2017-05-19 11:38:07

标签: python tkinter

from Tkinter import*
from tkMessageBox import*
import random
import time

n = 1
a = 1

class GameFrame(Frame):

    def __init__(self):
        global a
        global n
        Frame.__init__(self)
        self.master.title("Be a Billionaire")
        self.master.geometry("1200x700")
        self.master.resizable(0,0)
        self.grid()

        self._mission = Label(self, text = "Your Mission is to Earn 1,000,000,000$" ,font = ("Arial", 30, "bold"))
        self._mission.grid()

        counter1 = IntVar()
        counter1.set(0)

        self._currentmoney = Label(self, text = "Your Current Money:" ,font = ("Arial", 30, "bold"))
        self._currentmoney.grid()

        self._currentmoney = Label(self, textvariable = counter1 ,font = ("Arial", 30, "bold"))
        self._currentmoney.grid()

        ###########################################################Click###########################################################
        self._moneyearn = Button(self, text = "Click Me to Earn Money", font = ("Arial", 30, "bold"), command = lambda: increasemoney(a))
        self._moneyearn.grid()

        self._clickupgrade1 = Button(self, text = "Click Me to Upgrade Click", font = ("Arial", 15, "bold"), command = lambda: upgradeclick())
        self._clickupgrade1.grid()

        self._costlabel1 = Label(self, text = "Cost:", font = ("Arial", 15, "bold"))
        self._costlabel1.grid()

        counter2 = IntVar()
        counter2.set(2)
        self._neededmoney1 = Label(self, textvariable = counter2, font = ("Arial", 15, "bold"))
        self._neededmoney1.grid()

        def increasemoney(x):
            counter1.set(counter1.get() + x)

        def upgradeclick():
            global a
            global n
            if counter1.get() >= n*(n+1)*(n+2):
                a += 5*n
                counter1.set(counter1.get() - n*(n+1)*(n+3))
                counter2.set((n+1)*(n+2)*(n+3))
                n += 1               
            else:
                showwarning(message = "You Don't Have Enough Money!", parent = self)        
        ###########################################################################################################################

        ##########################################################Lottery##########################################################
        self._lotterytitle = Label(self, text = "LOTTERY! : You can win 0$ ~ 100*(money you put in)", font = ("Arial", 15, "bold"))
        self._lotterytitle.grid()

        self._lotterymoney = IntVar()
        self._lotteryentry = Entry(self, textvariable = self._lotterymoney, font = ("Arial", 15, "bold"))
        self._lotteryentry.grid()

        self._lotterybutton = Button(self, text = "See the results", font = ("Arial", 15, "bold"), command = lambda: lottery())
        self._lotterybutton.grid()

        def lottery():
            x = self._lotterymoney.get()
            if x <= counter1.get():
                l = random.randint(1, 100)
                if 100 >= l > 99:
                    counter1.set(counter1.get() + x*99)
                    showinfo(message = "Congratulations! You've won First Prize(100*(Money you have put in))", parent = self)
                elif 99 >= l > 94:
                    counter1.set(counter1.get() + x*29)
                    showinfo(message = "Congratulations! You've won Second Prize(20*(Money you have put in))", parent = self)
                elif 94 >= l > 80:
                    counter1.set(counter1.get() + x*9)
                    showinfo(message = "Congratulations! You've won Third Prize(10*(Money you have put in))", parent = self)
                else:
                    counter1.set(counter1.get() - x)
                    showinfo(message = "Sorry! You've Lost", parent = self)
            else:
                showwarning(message = "You Don't Have Enough Money!", parent = self)
        ###########################################################################################################################

        ########################################Additional Income(Earns Money Every Second)########################################
        self._additionaltitle = Label(self, text = "Additional Income(Earns Money Every Second)", font = ("Arial", 15, "bold"))
        self._additionaltitle.grid()




def main():
    GameFrame().mainloop()

main()

额外收入部分,我想将某种值 m 添加到counter1,但我不知道如何做到这一点。

1 个答案:

答案 0 :(得分:2)

要在tkinter应用中每秒运行一次函数,请使用after

  

之后(delay_ms,callback = None,* args)

     

注册在给定时间后调用的警报回调。

应该在要定期调用的函数的末尾调用此方法。例如,设想一个MyWidget类,我想每秒运行foo方法:

class MyWidget(tk.Widget):
    def foo(self):
        print("foo")
        self.after(1000, self.foo)
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