Question

我试图在我的游戏中创建商店，但是我遇到了pygame.mouse.get_pressed（）的问题。当用户点击按钮时，程序认为他们多次点击它，因此使用的信用额度超出了用户的预期。我想添加一个延迟，以便游戏不会这样做。我想点击一下来代表他们购买一个物品。我已经尝试降低该窗口的帧速率，但结果是一样的。这是我目前的代码。

这是发生所有鼠标操作的地方。

def button(x, y, w, h, ic, ac, action = None):
    global paused

    mouse = pygame.mouse.get_pos()
    click = pygame.mouse.get_pressed()

    if x + w > mouse[0] > x and y + h > mouse[1] > y:
        pygame.draw.rect(gameDisplay, ac, (x, y, w, h))

        if click[0] == 1 and action == Game:
            Game()
        if click[0] == 1 and action == quitgame:
            sys.exit()
        if click[0] == 1 and action == None:
            paused = False
        if click[0] == 1 and action == StartScreen:
            save()
            StartScreen()
        if click[0] == 1 and action == LootScreen:
            LootScreen()
        if click[0] == 1 and action == open_common_drop:
            open_common_drop()
        if click[0] == 1 and action == open_rare_drop:
            open_rare_drop()

    else:
        pygame.draw.rect(gameDisplay, ic, (x, y, w, h))

这是战利品店目前所在的地方。

def LootScreen():
    global current_drops

    loot = True

    while loot:
        for event in pygame.event.get():

            if event.type == pygame.QUIT:
                save()

                pygame.quit()
                sys.exit()

            if event.type == pygame.KEYDOWN:
                if event.key == pygame.K_t:
                    open_common_drop()
                elif event.key == pygame.K_y:
                    open_rare_drop()

                if event.key == pygame.K_ESCAPE:
                    StartScreen()

        gameDisplay.fill(gray)
        title('Loot Chests!')

        button(400, 150, 260, 50, blue, bright_blue, open_common_drop)
        button(695, 150, 260, 50, red, bright_red, open_rare_drop)
        button(display_width * 0.42, display_height / 1.15, 255, 50, red, bright_red, StartScreen)

        game_display_text('Open Common Chest (T)', 407, 165)
        game_display_text('Open Rare Chest (Y)', 725, 165)
        game_display_text('You Got: %s' % current_drops, 50, display_height / 2)
        game_display_text('Credits: %.2f' % player_loot_data['credit_count'], 15, 15)
        game_display_text('Main Menu', display_width * 0.47, display_height / 1.13)

        game_display_text('Janus\': %s' % player_loot_data['loot_data_one'] , 950, 500)
        game_display_text('Peace of Minds: %s' % player_loot_data['loot_data_two'], 950, 535)

        pygame.display.update()
        clock.tick(30)

Answer 1

我建议定时布尔开关。这是一个值得帮助的例子。

import time

boolswitch = False
timer = 0.0
clock = time.time()
x = time

while True:
    if timer > 0.25:
        boolswitch = True
        timer = 0.0
    else:
        x = time.time()
        timer += (x-clock)
        clock = x

    if boolswitch:
        click = pygame.mouse.get_pressed()
        if click[0] == 1:
            boolswitch = False
            clock = time.time()

Answer 2

您需要在按钮功能中使用bool开关。在这里，我以一种应该工作的方式重新处理了这个功能。

def button(x, y, w, h, ic, ac, action = None, held):
    global paused

    mouse = pygame.mouse.get_pos()
    click = pygame.mouse.get_pressed()

    if x + w > mouse[0] > x and y + h > mouse[1] > y:
        pygame.draw.rect(gameDisplay, ac, (x, y, w, h))

        if click[0] == 1:
            if held == False:
                if action == Game:
                    Game()
                elif action == quitgame:
                    sys.exit()
                elif action == None:
                    paused = False
                elif action == StartScreen:
                    save()
                    StartScreen()
                elif action == LootScreen:
                    LootScreen()
                elif action == open_common_drop:
                    open_common_drop()
                elif action == open_rare_drop:
                    open_rare_drop()
            held = True
        else:
            held = False

    else:
        pygame.draw.rect(gameDisplay, ic, (x, y, w, h))
    return held

如您所见，函数中添加了一个新变量held来处理按钮是否被保持。在此处保持并返回，因此每次调用该函数时都不会重置。

有了这个，让我告诉你为什么我这样写它，以及这个逻辑是如何工作的。

import pygame
#
display = pygame.display.set_mode((800,600))
clock = pygame.time.Clock()
#
held = False # Variable to handle if mouse button is held
#
RUNNING = True
while RUNNING:
    clock.tick(60)
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            RUNNING = False
    #
    button = pygame.mouse.get_pressed() # Get mouse state
    if button[0]: # Check if left mouse button is pressed
        if held == False: # Check if button is held down
            print(True) # If button is not held down, print true
        held = True # Set held eqaual to true for next iteration
    else: # If left mouse button is not pressed
        held = False # held is set to false, alowing event to happen again
    #
    display.fill((0,0,0))
    #
    pygame.display.flip()
#
pygame.quit()

以上是一个非常简单的独立程序，它还实现了保持变量以监视鼠标按钮是否被按住。在此程序中，变量held首先声明为False，因为未按下鼠标。然后，在主循环中，调用pygame.mouse.get_pressed()来获取鼠标输入，然后立即检查鼠标左键。如果按下鼠标左键，将检查held。如果为false，程序将打印True。然后将held设置为True以进行下一次迭代。如果未按住鼠标左键并将else:重置为其默认held状态，则会触发False语句。

如何在pygame中为鼠标点击添加延迟？

2 个答案: