如何自动关闭PySimpleGUI弹出窗口？

问题描述

我正在使用PySimpleGUI并打开自动关闭弹出窗口，但是即使按OK按钮，它也不会自动关闭。只有按'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，但不是。
也许是因为线程
请帮忙！

解决方法

对此有两种解决方案：

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 有关详细信息，请参阅参考：PySimpleGUI Reference - Popups

2. 您在事件循环中创建更新间隔。完整的循环应如下所示：

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

,

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

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()