如何点击鼠标

Question

我正在使用Pygame编写一个小游戏，我不确定如何点击鼠标。我已经弄清楚了一点，但说实话，我没有最好的时间。

这就是我所拥有的：

import pygame

pygame.init()

def menu():
    title_font = pygame.font.SysFont("monospace",64)
    button_font = pygame.font.SysFont("monospace", 30)
    screen_size = (1280,950)
    screen = pygame.display.set_mode(screen_size)

    background = (255,255,255)
    screen.fill(background)
    play_button = pygame.Rect(250,600,200,100)
    quit_button = pygame.Rect(850,600,200,100)
    controls_button = pygame.Rect(550,600,200,100)
    pygame.draw.rect(screen, (0,255,0), play_button)
    pygame.draw.rect(screen, (255,0,0), quit_button)
    pygame.draw.rect(screen, (255,229,0), controls_button)
    title = title_font.render("Fightastic!", 1, (0,0,0))
    screen.blit(title, (450,300))
    play_text = button_font.render("START",1,(0,0,0))
    screen.blit(play_text, (310,635))
    quit_text = button_font.render("QUIT",1,(0,0,0))
    screen.blit(quit_text, (910,635))
    controls_text = button_font.render("CONTROLS",1,(0,0,0))
    screen.blit(controls_text, (580,635))
    buttons = [play_button, quit_button, controls_button]

    while True:

        events = pygame.event.get()
        for event in events:
            if event.type == pygame.QUIT:
                exit()

        mouse_cursor = pygame.mouse.get_pos()
        mouse_pressed = pygame.mouse.get_pressed()

        option = 0
        for i in range(len(buttons)):
            if buttons[i].collidepoint( mouse_cursor):
                option = i+1

        if option == 1:
            print ("YO I GOT CLICKED")
        elif option == 2:
            print ("CLICKED MY DUDE")
        elif option == 3:
            quit()

    pygame.display.update()

menu()

游戏菜单是唯一需要点击的部分，所以我已经展示了。

谢谢！

Answer 1

您已经拥有pygame.mouse.get_pressed()，因此请使用它来检查点击了哪个按钮。

    option = 0

    if mouse_pressed[0]: # check if left button was clicked
        for i, but in enumerate(buttons, 1): # get button and its number
            if but.collidepoint(mouse_cursor): # check collision
                option = i # remember button number
                break      # no need to check other buttons

但是使用pygame.mouse.get_pressed()会产生一个问题 - 只要按住鼠标按钮就会返回True（所以看起来每秒都有很多次点击）所以如果你更改屏幕上的按钮和新按钮将在同一个地方，然后它会自动点击新按钮。更好地使用event.type == pygame.MOUSEBUTTONDOWN只创建一次点击。

while True:

    option = 0

    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            exit()
        if event.type == pygame.MOUSEBUTTONDOWN:
            for i, but in enumerate(buttons, 1):
                if but.collidepoint(event.pos):
                    option = i
                    break # no need to check other buttons

    if option == 1:
        print ("YO I GOT CLICKED")
    elif option == 2:
        print ("CLICKED MY DUDE")
    elif option == 3:
        quit()

    pygame.display.update()

<强>顺便说一句：

你可以保留带回调的按钮（没有()的功能名称）

    buttons = [
        (play_button, play_function),
        (controls_button, control_function),
        (quit_button, quit_function),
    ]

然后你可以直接调用function / callback（使用()）而不用option

            if event.type == pygame.MOUSEBUTTONDOWN:
                for rect, callback in buttons:
                    if rect.collidepoint(event.pos):
                        callback() # execute function
                        break # no need to check other buttons

完整的工作代码

import pygame

# --- constants --- (UPPER_CASE_NAMES)

SCREEN_WIDTH = 1280
SCREEN_HEIGHT = 950

WHITE = (255, 255, 255)

# --- functions --- (lower_case_names)

def play_function():
    print("YO I GOT CLICKED")

def controls_function():
    print("CLICKED MY DUDE")

def quit_function():
    pygame.quit()
    quit()

def menu(screen):

    # - init -

    title_font = pygame.font.SysFont("monospace", 64)
    button_font = pygame.font.SysFont("monospace", 30)

    # - objects -

    play_button = pygame.Rect(250,600,200,100)
    quit_button = pygame.Rect(850,600,200,100)
    controls_button = pygame.Rect(550,600,200,100)

    buttons = [
        (play_button, play_function),
        (controls_button, controls_function),
        (quit_button, quit_function),
    ]

    # - draws -

    screen.fill(WHITE)

    title = title_font.render("Fightastic!", 1, (0,0,0))
    screen.blit(title, (450,300))


    pygame.draw.rect(screen, (0,255,0), play_button)
    play_text = button_font.render("START",1,(0,0,0))
    screen.blit(play_text, (310,635))

    pygame.draw.rect(screen, (255,0,0), quit_button)
    quit_text = button_font.render("QUIT",1,(0,0,0))
    screen.blit(quit_text, (910,635))

    pygame.draw.rect(screen, (255,229,0), controls_button)
    controls_text = button_font.render("CONTROLS",1,(0,0,0))
    screen.blit(controls_text, (580,635))

    pygame.display.update()

    # - mainloop -

    clock = pygame.time.Clock()

    while True:

        for event in pygame.event.get():
            if event.type == pygame.QUIT:
                exit()
            if event.type == pygame.MOUSEBUTTONDOWN:
                for rect, callback in buttons:
                    if rect.collidepoint(event.pos):
                        callback() # execute function
                        break # no need to check other buttons

        clock.tick(5) # 5 FPS - to slow down game and use less CPU

# --- main ---

pygame.init()
screen = pygame.display.set_mode((SCREEN_WIDTH, SCREEN_HEIGHT))

menu(screen)

Answer 2

使用鼠标输入实际上相对容易使用，您可以用很少的代码编写强大的交互性。

我拿了你给的代码并做了一些调整和修正。

• 缩进行pygame.display.update()。

• 实施了完整的鼠标点击检测。

• 修改了option的使用方式和列表buttons的顺序，以便按钮按从左到右的顺序排列。

主要的补充是if语句中的mouse_pressed[0] == 1部分。这将获取鼠标按钮列表（左，右，中）和一个int 0或1，指示是否按下它们。因此，代码检查是否按下了左按钮。

以下是代码：

import pygame

pygame.init()

def menu():
    title_font = pygame.font.SysFont("monospace",64)
    button_font = pygame.font.SysFont("monospace", 30)
    screen_size = (1280,950)
    screen = pygame.display.set_mode(screen_size)

    background = (255,255,255)
    screen.fill(background)
    play_button = pygame.Rect(250,600,200,100)
    quit_button = pygame.Rect(850,600,200,100)
    controls_button = pygame.Rect(550,600,200,100)
    pygame.draw.rect(screen, (0,255,0), play_button)
    pygame.draw.rect(screen, (255,0,0), quit_button)
    pygame.draw.rect(screen, (255,229,0), controls_button)
    title = title_font.render("Fightastic!", 1, (0,0,0))
    screen.blit(title, (450,300))
    play_text = button_font.render("START",1,(0,0,0))
    screen.blit(play_text, (310,635))
    quit_text = button_font.render("QUIT",1,(0,0,0))
    screen.blit(quit_text, (910,635))
    controls_text = button_font.render("CONTROLS",1,(0,0,0))
    screen.blit(controls_text, (580,635))
    buttons = [play_button, controls_button, quit_button]

    while True:

        events = pygame.event.get()
        for event in events:
            if event.type == pygame.QUIT:
                exit()

        mouse_cursor = pygame.mouse.get_pos()
        mouse_pressed = pygame.mouse.get_pressed()

        option = -1

        for i in range(len(buttons)):
            if buttons[i].collidepoint( mouse_cursor ) and mouse_pressed[0] == 1:
                option = i

        if option == 0:
            print ("YO I GOT CLICKED")
        elif option == 1:
            print ("CLICKED MY DUDE")
        elif option == 2:
            quit()

        pygame.display.update()

menu()

我希望这有用，如果您有任何其他问题，请随时在下面发表评论！