python - time .sleep() 在命令中以错误的顺序发生；总是在函数的开头

Question

下面的代码是我正在开发的一个小型应用程序的精简版本(为了清晰起见)； child 拼写单词的应用程序。

问题

我遇到的问题是在函数flash_ Correct()中；它的目的是显示一个单词 5 秒，然后再次隐藏。

我一定有一个愚蠢的盲点，但无论我把time.sleep(5)放在哪里，函数都会以5秒的休息时间开始，而条目:self .entry2 从未出现:

但是，如果没有time.sleep(5)，它会正确显示:

我的盲点在哪里？

(精简的)代码:

#!/usr/bin/env python3
from gi.repository import Gtk, Pango, Gdk
import subprocess
import time

       
class InterFace(Gtk.Window):

    def __init__(self):

        Gtk.Window.__init__(self, title="Woorden raden")
        maingrid = Gtk.Grid()
        self.add(maingrid)
        maingrid.set_border_width(10)

        self.entry2 = Gtk.Entry()
        self.entry2.set_size_request(500,60)
        self.entry2.set_child_visible(False)
        self.entry2.modify_font(Pango.FontDescription('Ubuntu 30'))
        maingrid.attach(self.entry2, 0, 4, 4, 1)

        quitbutton = Gtk.Button("Stop", use_underline=True)
        quitbutton.modify_font(Pango.FontDescription('Ubuntu 20'))
        quitbutton.connect("clicked", self.on_close_clicked)
        maingrid.attach(quitbutton, 3, 7, 1, 1)

        showword_button = Gtk.Button("↺", use_underline=True)
        showword_button.modify_font(Pango.FontDescription('Ubuntu 25'))
        showword_button.connect("clicked", self.flash_correct)
        showword_button.set_size_request(60,20)
        maingrid.attach(showword_button, 0, 6, 1, 1)

    def flash_correct(self, button):
        # the time.sleep(5) seems to take place at the beginning
        # no matter in which order I set the commands
        self.entry2.set_text("Monkey")
        self.entry2.set_child_visible(True)
        time.sleep(5)
        self.entry2.set_child_visible(False)

    def on_close_clicked(self, button):
        Gtk.main_quit()


window = InterFace()
window.connect("delete-event", Gtk.main_quit)
window.set_default_size(330, 330)
window.set_resizable(False)
window.show_all()
Gtk.main()

Answer 1

您可以使用 time.time 在循环中调用 Gtk.main_iteration() 隐藏大约 5 秒，以避免您的应用变得无响应。

def hide(self, time_lapse):
    start = time.time()
    end = start + time_lapse
    while end > time.time():
        Gtk.main_iteration()

def flash_correct(self, button):
    # the time.sleep(5) seems to take place at the beginning
    # no matter in which order I set the commands
    self.entry2.set_text("Monkey")
    self.entry2.set_child_visible(True)
    self.hide(5)
    self.entry2.set_child_visible(False)

pygtk faq 7中有很好的解释。How can I force updates to the application windows during a long callback or other internal operation?

If you have a long-running callback or internal operation that tries to modify the application windows incrementally during its execution, you will notice that this doesn't happen; the windows of your app freeze for the duration.

This is by design: all gtk events (including window refreshing and updates) are handled in the mainloop, and while your application or callback code is running the mainloop can't handle window update events. Therefore nothing will happen in the application windows.

The trick here is to realize where your operation can take a while to return, or where it is dynamically changing the window contents, and add a code fragment like this wherever you want an update forced out:

while gtk.events_pending():
   gtk.main_iteration(False)

This tells gtk to process any window events that have been left pending. If your handler has a long loop, for instance, inserting this snippet as part of the loop will avoid it hanging the window till the callback has finished.

More eloquently, in the words of the great Malcolm Tredinnick, 'this requires using what should be called "Secret Technique #1 For Making Your Application Look Responsive"(tm):

添加 while gtk.events_pending(): 也可能没有什么坏处。