PySimpleGUI 冻结

Question

我在 python 3.8.1 中使用 PySimpleGui 4.15.1 编写了一个游戏冷拳击模拟器 2 的破解代码，如果我尝试单击 start/stop 按钮，gui冻结，但与 Start/Stop 按钮连接的脚本仍在运行。但是，如果我将连接到 Start/Stop 按钮的脚本更改为 Print('Chicken') 之类的东西，那么它就完全完美了⁫⁯⁫⁯⁫⁯⁫⁫⁫

代码如下：

import PySimpleGUI as sg
import ctypes
import time
import pynput

sg.theme('DarkBrown1')

layout = [  [sg.Text('BOXING SIMULATOR 2 HACK', size=(20, 2), justification='center')],
            [sg.Text('', size=(10, 2), font=('Helvetica', 20), justification='center', key='_OUTPUT_')],
            [sg.T(' ' * 5), sg.Button('Start/Stop', focus=True), sg.Quit()]]

window = sg.Window('BOXING SIMULATOR 2 HACK', layout)

SendInput = ctypes.windll.user32.SendInput

PUL = ctypes.POINTER(ctypes.c_ulong)
class KeyBdInput(ctypes.Structure):
    _fields_ = [("wVk", ctypes.c_ushort),
                ("wScan", ctypes.c_ushort),
                ("dwFlags", ctypes.c_ulong),
                ("time", ctypes.c_ulong),
                ("dwExtraInfo", PUL)]

class HardwareInput(ctypes.Structure):
    _fields_ = [("uMsg", ctypes.c_ulong),
                ("wParamL", ctypes.c_short),
                ("wParamH", ctypes.c_ushort)]

class MouseInput(ctypes.Structure):
    _fields_ = [("dx", ctypes.c_long),
                ("dy", ctypes.c_long),
                ("mouseData", ctypes.c_ulong),
                ("dwFlags", ctypes.c_ulong),
                ("time",ctypes.c_ulong),
                ("dwExtraInfo", PUL)]

class Input_I(ctypes.Union):
    _fields_ = [("ki", KeyBdInput),
                 ("mi", MouseInput),
                 ("hi", HardwareInput)]

class Input(ctypes.Structure):
    _fields_ = [("type", ctypes.c_ulong),
                ("ii", Input_I)]


def PressKeyPynput(hexKeyCode):
    extra = ctypes.c_ulong(0)
    ii_ = pynput._util.win32.INPUT_union()
    ii_.ki = pynput._util.win32.KEYBDINPUT(0, hexKeyCode, 0x0008, 0, ctypes.cast(ctypes.pointer(extra), ctypes.c_void_p))
    x = pynput._util.win32.INPUT(ctypes.c_ulong(1), ii_)
    SendInput(1, ctypes.pointer(x), ctypes.sizeof(x))

def ReleaseKeyPynput(hexKeyCode):
    extra = ctypes.c_ulong(0)
    ii_ = pynput._util.win32.INPUT_union()
    ii_.ki = pynput._util.win32.KEYBDINPUT(0, hexKeyCode, 0x0008 | 0x0002, 0, ctypes.cast(ctypes.pointer(extra), ctypes.c_void_p))
    x = pynput._util.win32.INPUT(ctypes.c_ulong(1), ii_)
    SendInput(1, ctypes.pointer(x), ctypes.sizeof(x))


HACK_running, counter = False, 0

while True:                                 
    event, values = window.read(timeout=10) 
    if event in (None, 'Quit'):            
        break
    elif event == 'Start/Stop':
        HACK_running = not HACK_running
    if HACK_running:
            PressKeyPynput(0x02)
            time.sleep(0.08)
            ReleaseKeyPynput(0x02)
            PressKeyPynput(0x11)
            time.sleep(0.5)
            ReleaseKeyPynput(0x11)
            PressKeyPynput(0x1F)
            time.sleep(0.6)
            ReleaseKeyPynput(0x1F)
            PressKeyPynput(0x02)
            time.sleep(0.08)
            ReleaseKeyPynput(0x02)
            time.sleep(300)

Answer 1

如果没有某种刷新或读取操作，您无法在 GUI 中执行很长的运行任务。

我回答了上一个 post 与这个相同的问题，指出问题是永远不会退出的内部 while 循环。

如果您必须让内部循环永远执行，那么您至少需要刷新 GUI。您可以通过调用 window.refresh 来完成此操作，如下所示。

while True:                                 
    event, values = window.read(timeout=10) 
    if event in (None, 'Quit'):            
        break
    elif event == 'Start/Stop':
        HACK_running = not HACK_running
    if HACK_running:
            PressKeyPynput(0x02)
            time.sleep(0.08)
            ReleaseKeyPynput(0x02)
            PressKeyPynput(0x11)
            time.sleep(0.5)
            ReleaseKeyPynput(0x11)
            PressKeyPynput(0x1F)
            time.sleep(0.6)
            ReleaseKeyPynput(0x1F)
            PressKeyPynput(0x02)
            time.sleep(0.08)
            ReleaseKeyPynput(0x02)
            time.sleep(300)
            window.refresh()

你永远不应该在 PySimpleGUI 事件循环中休眠。

更好的方法是调用 window.read(timeout=400) 而不是 time.sleep(0.4)。这将使 OS 不会认为您的程序已挂起。

PySimpleGUI 冻结

PySimpleGUI Freezes

python

python-3.x

pynput

pysimplegui