如何自动关闭 PySimpleGUI 弹出窗口？

Question

我正在使用 PySimpleGUI 并打开自动关闭弹出窗口，但它不会自行关闭，即使我按下 OK 按钮也是如此。仅当我按“X”
这是我的代码：

import PySimpleGUI as sg
import threading
import time

layout = [
    [sg.Text('', size=(40, 1))],
    [sg.Text('', size=(30, 2)), sg.Text('Press "Start" button', size=(55, 12), key='-MAIN-')],
    [sg.Button('Start', size=(10,2))],
]
window = sg.Window('APP', layout)
while True:
    event, values = window.read()
    if event == sg.WIN_CLOSED:
        break
    if event == 'Start':
        def thread_reminder(seconds):
            seconds = 0
            while True:
                seconds += 1
                time.sleep(1)
                print(seconds)
                if seconds == 10:
                    sg.popup_auto_close("1 minute passed")
        threading.Thread(target=thread_reminder, args=(1,), daemon=True).start()
window.close()

这给了我这个错误或异常，我不知道：

Exception in thread Thread-1:
Traceback (most recent call last):
  File "C:\Users\User\AppData\Local\Programs\Python\Python38\lib\threading.py", line 932, in _bootstrap_inner
    self.run()
  File "C:\Users\User\AppData\Local\Programs\Python\Python38\lib\threading.py", line 870, in run
    self._target(*self._args, **self._kwargs)
  File "C:/Users/User/Downloads/jkl;'.py", line 23, in thread_reminder
    sg.popup_auto_close("1 minute passed")
  File "C:\Users\User\AppData\Local\Programs\Python\Python38\lib\site-packages\PySimpleGUI\PySimpleGUI.py", line 15782, in PopupAutoClose
    return Popup(*args, title=title, button_color=button_color, background_color=background_color, text_color=text_color,
  File "C:\Users\User\AppData\Local\Programs\Python\Python38\lib\site-packages\PySimpleGUI\PySimpleGUI.py", line 15353, in Popup
    button, values = window.read()
  File "C:\Users\User\AppData\Local\Programs\Python\Python38\lib\site-packages\PySimpleGUI\PySimpleGUI.py", line 7568, in Read
    results = self._read(timeout=timeout, timeout_key=timeout_key)
  File "C:\Users\User\AppData\Local\Programs\Python\Python38\lib\site-packages\PySimpleGUI\PySimpleGUI.py", line 7623, in _read
    self._Show()
  File "C:\Users\User\AppData\Local\Programs\Python\Python38\lib\site-packages\PySimpleGUI\PySimpleGUI.py", line 7395, in _Show
    StartupTK(self)
  File "C:\Users\User\AppData\Local\Programs\Python\Python38\lib\site-packages\PySimpleGUI\PySimpleGUI.py", line 12817, in StartupTK
    window.TKroot.mainloop()
  File "C:\Users\User\AppData\Local\Programs\Python\Python38\lib\tkinter\__init__.py", line 1420, in mainloop
    self.tk.mainloop(n)
RuntimeError: Calling Tcl from different apartment
*** Faking timeout ***

*** Faking timeout *** 表示必须关闭弹出窗口，但没有。
也许是因为线程
请帮忙！

Answer 1

不要在另一个线程中调用 PySimpleGUI/tkinter 的 GUI 方法。 使用window.write_event_value 生成一个事件并在事件循环中处理它。

修改后的代码

import PySimpleGUI as sg
import threading
import time

def thread_reminder(seconds, window):
    count = 0
    while count < seconds :
        count += 1
        time.sleep(1)
        print(count)
    window.write_event_value('Alarm', "1 minute passed")

layout = [
    [sg.Text('', size=(40, 1))],
    [sg.Text('', size=(30, 2)), sg.Text('Press "Start" button', size=(55, 12), key='-MAIN-')],
    [sg.Button('Start', size=(10,2))],
]
window = sg.Window('APP', layout)

while True:
    event, values = window.read()
    if event == sg.WIN_CLOSED:
        break
    elif event == 'Start':
        threading.Thread(target=thread_reminder, args=(10, window), daemon=True).start()
    elif event == 'Alarm':
        message = values[event]
        sg.popup_auto_close(message)

window.close()

Answer 2

有两种解决方案：

1. PySimpleGUI 提供在给定时间后关闭的弹出窗口：sg.popup_timed('popup_timed') # Automatically closes sg.popup_auto_close('popup_auto_close') # Same as PopupTimed 详见参考：PySimpleGUI Reference - Popups

2. 您在事件循环中创建更新间隔。完整的循环应该是这样的：

while True:
    event, values = window.Read(timeout = 1000 * 10)  # in milliseconds

    if event in ('__TIMEOUT__',):
        print('timed execution inside event loop')
        ### put you window closing commands here ###

    if event in (sg.WIN_CLOSED,): break