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Asked:2022-08-02 04:05:10 +0000 UTC2022-08-02 04:05:10 +0000 UTC

如何在不阻塞 GUI 的情况下循环超时？

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有一段代码将未启动的沙箱标记为红色。该函数start_btn()检查 CheckButtons，然后启动标记的按钮，标签立即着色为绿色。问题是什么 - 如何确保 self.builder.get_object(f"status_label_{box}").set_markup("бла бла бла") 在重新调用start_box()循环启动框的函数之前，在该行之后有 N 秒的超时，但是这个超时不应该阻塞 GUI，也就是说，它会time.sleep()飞走立即地

import gi
import os

gi.require_version('Gtk', '3.0')
from gi.repository import Gtk, Gdk, GLib

sandboxie_path = "C:/Program Files/Sandboxie-Plus/Start.exe"  # Путь к Start.exe
Steam_path = "C:/Program Files (x86)/Steam/steam.exe"  # Путь к steam.exe
tn_host = "127.0.0.1"
port_list = ["2121", "2122", "2123", "2124", "2125", "2126"]
server_list = ["85.88.162.140:27031"]
check_list = []


def start_box(box_num, tn_port, server, sandbox_path, steam_path):
    os.popen(f'"{sandbox_path}" /box:{box_num} "{steam_path}" -applaunch 730 -netconport {tn_port} -novid '
             f'-window -nosound +fps_max 30 -w 640 -h 480 +left connect {server}')


class Main:

    def __init__(self):
        gladeFile = "res/csgo.glade"
        self.builder = Gtk.Builder()
        self.builder.add_from_file(gladeFile)
        self.builder.connect_signals(self)
        window = self.builder.get_object("main")
        window.connect('delete-event', Gtk.main_quit)
        window.show()

    def red_mark(self):
        for num in range(1, 7):
            self.builder.get_object(f"status_label_{num}").set_markup("<span background='#FA6DA4' "
                                                                      "foreground='#000000'>OFF</span>")

    def start_btn(self, widget):
        for box in range(1, 7):
            if self.builder.get_object(f"check_btn_{box}").get_active():
                start_box(box, port_list[box - 1], server_list[0], sandboxie_path, Steam_path)
                self.builder.get_object(f"status_label_{box}").set_markup("<span background='#76EE98' "
                                                                          "foreground='#000000'>ON</span>")
                # Тайм-аут на N секунд


if __name__ == '__main__':
    main = Main()
    main.red_mark()
    Gtk.main()

所有盒子 两者都推出

python-3.x gui

1 个回答

  1. Best Answer

    决定这样：

    我将标签绘画翻译self.builder.get_object(f"status_label_{box_num}").set_markup("<span background='#76EE98' foreground='#000000'>ON</span>")成一个函数start_box()，在函数中的循环中start_btn()添加bool了一个带初始值的计数器，False并将循环分为两个条件，让True遇到的第一个将计数器切换到True并立即启动第一个选中的框，接下来的间隔为 15 秒，使用threading.Timer.

    这是目前的样子

        def start_btn(self, widget):
        counter = False
        for box in range(1, 7):
            if self.builder.get_object(f"check_btn_{box}").get_active() and not counter:
                counter = True
                self.start_box(box, port_list[box - 1], server_list[0], sandboxie_path, Steam_path)
                # self.builder.get_object(f"status_label_{box}").set_markup("<span background='#76EE98' "
                #                                                           "foreground='#000000'>ON</span>")
            elif self.builder.get_object(f"check_btn_{box}").get_active() and counter:
                threading.Timer(15, self.start_box,
                                (box, port_list[box - 1], server_list[0], sandboxie_path, Steam_path)).start()
    • 0

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