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[英]`pygame.mouse.get_pressed()` reports clicks when mouse is not clicked
[英]How to get mouse clicks
我正在用Pygame编写一个小游戏,我不确定如何获得鼠标点击。 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()
游戏的菜单是唯一需要点击的部分,这就是我所展示的全部内容。
谢谢!
你已经有pygame.mouse.get_pressed()
所以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 - 它会一直按住鼠标按钮返回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,新按钮就会一样然后它会自动点击新按钮。 最好使用event.type == pygame.MOUSEBUTTONDOWN
,它只创建一次点击。
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()
BTW:
您可以保留带回调的按钮(函数名称不带()
)
buttons = [
(play_button, play_function),
(controls_button, control_function),
(quit_button, quit_function),
]
然后你可以直接调用function / callback(using ()
)而无需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)
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复。
缩进行pygame.display.update()
。
实现了完整的鼠标点击检测。
修改了使用option
的方式和列表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。
这是代码:
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!
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