Question

下面的代码是我试图在python上创建的数学测验游戏的“添加”部分。每当点击button4（使用submit_answer命令）时，倒计时秒似乎比以前快两倍。有没有办法让每次点击按钮4时倒计时时间长度保持不变？

class Add(tk.Frame):

    def __init__(self, parent, controller):
        tk.Frame.__init__(self, parent, background="blue")
        button = tk.Button(self, text="Main Menu", font=LARGE_FONT, background="white",
                           command=lambda: controller.show_frame(Main))
        button.place(relx=0.83, rely=0.92)

        self.button4 = tk.Button(self, text="Start Game", font=LARGE_FONT, background="white",
                           command=self.time_start)
        self.button4.place(relx=0.4, rely=0.8, relheight=0.1, relwidth=0.2)

    def time_start(self):
        self.button4.destroy()

        self.x = int(random.uniform(1,10))
        self.y = int(random.uniform(50,100))
        self.z = int(random.uniform(10,50))
        print(self.x+self.y+self.z)

        self.qlabel = tk.Label(self, text= (self.x,"+",self.y,"+",self.z,"=",), font=WORD_FONT, bg="blue")
        self.qlabel.place(relx=0.25, rely=0.3, relheight=0.05, relwidth=0.3)

        self.e1 = tk.Entry(self)
        self.e1.place(relx=0.5, rely=0.3, relheight=0.05, relwidth=0.2)

        self.ebutton = tk.Button(self, text="Done", font=TITLE_FONT, background="white",
                            command=self.submit_answer)
        self.ebutton.place(relx=0.35, rely=0.5, relheight=0.2, relwidth=0.3)

        self.label = tk.Label(self, text="", width=10, font=LARGE_FONT, bg="white")
        self.label.place(relx=0.1, rely=0.1)
        self.remaining = 0
        self.countdown(10)

    def submit_answer(self):
        a = (self.x+self.y+self.z)
        if int(self.e1.get()) == a:
            answerlabel = tk.Label(self, text="Correct!", font=LARGE_FONT, background="blue")
            answerlabel.pack()
            self.label.destroy()
            self.qlabel.destroy()
            self.e1.destroy()
            self.ebutton.destroy()

            self.time_start()
        else:
            answerlabel = tk.Label(self, text="Game Over.", font=LARGE_FONT, background="blue")
            answerlabel.pack()
            self.ebutton.destroy()

    def countdown(self, remaining = None):
        if remaining is not None:
            self.remaining = remaining
        if self.remaining <= 0:
            self.label.configure(text="Time's up!")
            self.ebutton.destroy()           
        else:
            self.label.configure(text="%d" % self.remaining)
            self.remaining = self.remaining - 1
            self.after(1000, self.countdown)

Answer 1

调用after会返回一个ID，您可以将其传递给after_cancel以删除待处理的作业。因此，在开始新的计时之前，请跟踪此ID并杀死任何先前的计时器。

class Add(tk.Frame):
    def __init__(self, parent, controller):
        ...
        # initialize after_id to none
        self.after_id = None
        ...

def time_start(self):
    # kill the old timer
    if self.after_id is not None:
        self.after_cancel(self.after_id)
    ...
...
def countdown(self, remaining = None):
    ...
    if self.remaining <= 0:
        ...  
        self.after_id = None         
    else:
        ...
        self.after_id = self.after(1000, self.countdown)

如何使用按钮（tkinter）重置python中的倒数计时器

1 个答案: