阻止方法调用 n 秒，显示剩余时间的消息

Question

我正在开发简单的 GUI，但我被困住了。这是基本流程：

1. 显示文本text 和：
   • 节省时间time_pressed
   • 启动progressbar 并更新它，直到time_notallow 过期。
2. 如果用户按下<Next>，查看time_notallow 中指定的秒数是否已过， 如果没有，则不允许显示下一个text。

基本上，我想阻止用户调用绑定到<Right> 键的方法，直到time_notallow 通过，并显示一个进度条来通知他们需要等待多长时间。由于我用的是bind，比如……

    self.master.bind('<Right>', self.text_next)

...我没有.after()，就像在小部件中一样。

我尝试过的

1. master.after() 将bind 设置为None 并在time_notallow 之后将bind 设置为self.text_next，但它不起作用。
2. 创建了一个thread，它与while True 循环，以不断检查time_notallow 是否通过，但应用程序崩溃。

任何帮助表示赞赏。

编辑：一个解决方案。在 .after 中使用 lambda() 来计算秒数（感谢 Bryan Oakley）

"""

Stripped-down version to figue out time/event/threading stuff.

What this has to do:

1. Show the window and some text.
2. User presses Next arrow and new text shows. Paint the label red.
3. Prevent user form pressing again (unbind all keys), until 2 seconds passed 
   (time_wait).
4. Make notice of passed time and after 2 seconds bind the keys again and
   paint the label green.
5. Loop the steps 2-4.

"""

import sys
import tkinter as tk
from tkinter import W, E, S, N

class Test(tk.Frame):

    def __init__(self, master=None):
        """Draw the GUI"""
        tk.Frame.__init__(self, master)
        self.draw_widgets()
        self.grid()
        self.time_wait = 2
        self.locked = False
        self.bind_keys()
        self.counter = 0

    def draw_widgets(self):
        """Draw all the widgets on the frame."""
        text = 'Just a sample sentence.'
        #Label with the sentence
        self.lbl_text = tk.Label(self, anchor="center", relief='groove')
        self.lbl_text['text'] = text
        self.lbl_text['font'] = ('Helvetica', 27)
        self.lbl_text.grid(column=0, row=0, sticky=W+E+S+N)
        self.lbl_note = tk.Label(self, anchor="center", relief='groove',
            bg='green')
        self.lbl_note.grid(column=0, row=1, sticky=W+E+S+N)

    def text_next(self, event):
        """Get next text"""
        if not self.locked:
            self.counter += 1
            self.lbl_text['text'] = 'The text number %s!' % self.counter
            self.bind_tonone()
            self.lock(self.time_wait)

    def text_previous(self, event):
        """Get previous text"""
        if not self.locked:
            self.counter -= 1
            self.lbl_text['text'] = 'The text number %s!' % self.counter
            self.bind_tonone()
            self.lock(self.time_wait)

    def bind_keys(self):
        """Bind the keys"""
        self.master.bind('<Left>', self.text_previous)
        self.master.bind('<Right>', self.text_next)
        self.master.bind('<Escape>', self.exit)
        self.lbl_note['bg'] = 'green'

    def bind_tonone(self):
        """Unbind the keys"""
        self.master.bind('<Left>', None)
        self.master.bind('<Right>', None)
        self.master.bind('<Escape>', None)
        self.lbl_note['bg'] = 'red'

    def lock(self, n):
        if n == 0:
            self.locked = False
            self.lbl_note['text'] = ''
            self.lbl_note['bg'] = 'green'

        else:
            self.locked = True
            self.lbl_note['text'] = 'Locked for %s more seconds' % n
            self.lbl_note.after(1000, lambda n = n - 1: self.lock(n))

    def exit(self, event):
        """Exit the program."""
        sys.exit()

def start():
    """Start the gui part."""
    root = tk.Tk()
    app = Test(master=root)
    app.mainloop()

if __name__ == '__main__':
    start()

Answer 1

您不需要线程或计时器来解决此问题。您所需要的只是一个需要几秒钟等待的过程，并让它每秒调用一次，直到数字降至零。

它看起来像这样（在我的脑海中，未经测试）：

def __init__(...):
    ...
    self.locked = False
    ...

def text_next(self, event):
    if not self.locked:
        <do the "next" logic>
        self.lock(10) # lock for 10 seconds

def text_previous(self, event):
    if not self.locked:
        <do the "previous" logic>
        self.lock(10) # lock for 10 seconds

def lock(self, n):
    if n == 0:
        self.locked = False
        self.status.config(text="")
    else:
        self.locked = True
        self.status.config(text="Locked for %s more seconds" % n)
        self.status.after(1000, lambda n=n-1: self.lock(n))

Answer 2

尝试使用专门设计的threading.Timer 对象而不是自旋等待线程怎么样？

Answer 3

使用计时器的快速修复方法是进行一些重构。最初将您的 self.track 设置为 None，然后仅将其设置为在箭头上运行/阻止。

import sys
import threading
import Tkinter as tk
from Tkinter import W, E, S, N

class Test(tk.Frame):

    def __init__(self, master=None):
        """Draw the GUI"""
        tk.Frame.__init__(self, master)
        self.draw_widgets()
        self.grid()
        # Track wait times
        self.time_wait = 2
        # Timer
        self.track = None
        self.bind_keys()
        self.counter = 0

    def draw_widgets(self):
        """Draw all the widgets on the frame."""
        text = 'Just a sample sentence.'
        #Label with the sentence
        self.lbl_text = tk.Label(self, anchor="center", relief='groove')
        self.lbl_text['text'] = text
        self.lbl_text['font'] = ('Helvetica', 27)
        self.lbl_text.grid(column=0, row=0, sticky=W+E+S+N)
        self.lbl_note = tk.Label(self, anchor="center", relief='groove',
            bg='green')
        self.lbl_note.grid(column=0, row=1, sticky=W+E+S+N)

    def text_next(self, event):
        """Get next text"""
        if not self.track or not self.track.is_alive():
            self.track = threading.Timer(self.time_wait, self.bind_keys)
            self.counter += 1
            self.lbl_text['text'] = 'The text number %s!' % self.counter
            self.bind_tonone()
            self.track.start()

    def text_previous(self, event):
        """Get previous text"""
        self.counter -= 1
        self.lbl_text['text'] = 'The text number %s!' % self.counter

    def bind_keys(self):
        """Bind the keys"""
        self.master.bind('<Left>', self.text_previous)
        self.master.bind('<Right>', self.text_next)
        self.master.bind('<Escape>', self.exit)
        self.lbl_note['bg'] = 'green'

    def bind_tonone(self):
        """Unbind the keys"""
        self.master.bind('<Left>', None)
        self.master.bind('<Right>', None)
        self.master.bind('<Escape>', None)
        self.lbl_note['bg'] = 'red'

    def exit(self, event):
        """Exit the program."""
        sys.exit()

def start():
    """Start the gui part."""
    root = tk.Tk()
    app = Test(master=root)
    app.mainloop()

if __name__ == '__main__':
    start()

不过，在预览版中，@Bryan Oakley 有一个更好的解决方案。