python - 如何删除 pygbutton 创建的按钮

Question

我想删除 PygButton 创建的按钮。我是这样创建的:

button1 = pygbutton.PygButton((50, 50, 60, 30), '1')
button2 = pygbutton.PygButton((120, 50, 60, 30), '2')
button3 = pygbutton.PygButton((190, 50, 60, 30), '3')
allButtons = (button1,button2,button3)

for b in allButtons:
    b.draw(screen)

但是，一旦单击按钮，我想清除屏幕中的按钮并在屏幕上显示其他内容。

我该怎么做？

Answer 1

我想到的总体想法是在按下按钮后创建一个新屏幕。

基本上，我有一个名为 buttonhasbeenpressed 的 bool 值。在按下按钮之前，我们只是检查事件是否是按钮按下。按下后，我们将 bool 值设置为 True，“清除”背景(通过在旧屏幕上创建新屏幕)，然后继续执行我们想做的任何其他操作。我的示例代码仅“删除”按钮，更改背景颜色，并更改窗口上的标题，但您可以使用此想法来更改游戏按钮按下后的状态。

这是示例，您应该能够在您的计算机上运行该示例来进行测试。

import pygame,pygbutton
from pygame.locals import *
pygame.init()

#Create the "Pre Button Press Screen"
width = 1024
height = 768
screen = pygame.display.set_mode([width,height])
pygame.display.set_caption('OLD SCREEN NAME')
background = pygame.Surface(screen.get_size())
background = background.convert()
background.fill((250, 250, 250))
screen.blit(background, [0,0])
pygame.display.flip()
button1 = pygbutton.PygButton((50, 50, 60, 30), '1')
button2 = pygbutton.PygButton((120, 50, 60, 30), '2')
button3 = pygbutton.PygButton((190, 50, 60, 30), '3')
buttonhasbeenpressed = False

def screenPostButtonPress():
    width = 1024
    height = 768
    screen = pygame.display.set_mode([width,height])
    pygame.display.set_caption('NEW SCREEN NAME!!!!!!!')
    background = pygame.Surface(screen.get_size())
    background = background.convert()
    background.fill((20, 20, 40))
    screen.blit(background, [0,0])
    pygame.display.flip()
    #buttons not on screen after a button has been pressed

def waitingForButtonClick():
    allButtons = [button1,button2,button3]        
    buttonevent1 = button1.handleEvent(event)
    buttonevent2 = button2.handleEvent(event)
    buttonevent3 = button3.handleEvent(event)

    for b in allButtons:
        b.draw(screen)

    if 'click' in buttonevent1 or 'click' in buttonevent2 or 'click' in buttonevent3:
        return False
    return True

while True:
    for event in pygame.event.get():
        if event.type == QUIT:
            pygame.quit()
            sys.exit()
    #Wait for a button to be pressed, once one has, "clear" the screen by creating a new screen
    if buttonhasbeenpressed == False and waitingForButtonClick() == False:
       buttonhasbeenpressed = True
       screenPostButtonPress()
    pygame.display.update()