Python是我的第一语言,因此如果没有意义,请理解我。我正在用python制作虚拟宠物。我将Tkinter用于GUI。我想每5秒在Action类中运行一次display方法。但是,如果我使用线程,时间表或时间并将代码放在按钮的正前方,它就会冻结并只运行显示方法。如何在不中断按钮和其他部件的情况下每5秒运行一次显示方法?
class Pet:
def __init__(self, name, hunger=0, boredom=0, tiredness=0, sickness=False,
age=0, waste=0):
self.__name = name
self.hunger = hunger
self.boredom = boredom
self.tiredness = tiredness
self.sickness = sickness
self.age = age
self.waste = waste
self.choice = choice
class Action(Pet):
def __init__(self, name):
Pet.__init__(self, name, hunger=0, boredom=0, tiredness=0,
sickness=False, age=0, waste=0)
self.name = name
def display(self):
print("------------")
print("hunger", self.hunger)
print("boredom", self.boredom)
print("tiredness", self.tiredness)
print("sickness", self.sickness)
print("age", self.age)
class Window(Frame):
def openmenu(self):
petname = self.petnameEntry.get()
user_pet = Action(petname)
print("I am your pet,", petname)
window = tk.Toplevel(root)
w = Label(window, text="What would you like to do?")
w.pack()
btFeed = Button(window, text="Feed", command=lambda: user_pet.eat())
btFeed.pack(pady=3)
btPlay = Button(window, text="Play", command=lambda:
user_pet.play())
btPlay.pack(pady=3)
python参考方案
您可以使用After
方法。
import tkinter as tk
root = tk.Tk()
class Pet:
def __init__(self, name, hunger=0, boredom=0, tiredness=0, sickness=False,
age=0, waste=0):
self.__name = name
self.hunger = hunger
self.boredom = boredom
self.tiredness = tiredness
self.sickness = sickness
self.age = age
self.waste = waste
class Action(Pet):
def __init__(self, name):
Pet.__init__(self, name, hunger=0, boredom=0, tiredness=0,
sickness=False, age=0, waste=0)
self.name = name
def display(self):
print("------------")
print("hunger", self.hunger)
print("boredom", self.boredom)
print("tiredness", self.tiredness)
print("sickness", self.sickness)
print("age", self.age)
root.after(5000,self.display)
root.after(5000,Action(Pet).display)
root.mainloop()
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