我正在尝试使用Pygame制作一个简单的菜单,但我发现每当我使用pygame.mouse.get_position时,它确实是我想要的但是我必须继续移动鼠标以使我的图片保持blitting。
import pygame
import sys
pygame.init()
screen = pygame.display.set_mode((800,600))
pygame.display.set_caption('cursor test')
cursorPng = pygame.image.load('resources/images/cursor.png')
start = pygame.image.load('resources/images/menuStart.jpg')
enemy = pygame.image.load('resources/images/enemy-1.png')
white = (255,255,255)
black = (0,0,0)
clock = pygame.time.Clock()
FPS = 60
while True:
screen.fill(white)
pygame.mouse.set_visible(False)
x,y = pygame.mouse.get_pos()
x = x - cursorPng.get_width()/2
y = y - cursorPng.get_height()/2
screen.blit(cursorPng,(x,y))
for event in pygame.event.get():
if event.type == pygame.QUIT:
pygame.quit()
sys.exit()
elif event.type == pygame.KEYDOWN:
if event.key == pygame.K_ESCAPE:
pygame.quit()
sys.exit()
elif event.type == pygame.MOUSEMOTION:
if x < 50 and y < 250:
screen.blit(enemy,(100,100))
clock.tick(FPS)
pygame.display.update()
出了什么问题?
答案 0 :(得分:0)
看看你的代码:
for event in pygame.event.get():
...
elif event.type == pygame.MOUSEMOTION:
if x < 50 and y < 250:
screen.blit(enemy,(100,100))
您检查事件,如果您检测到鼠标正在被移动(并且只有那时),则将图像绘制到屏幕上。
如果您想要在不移动鼠标的情况下绘制图像,只需停止检查MOUSEMOTION
事件并简单地绘制图像:
while True:
screen.fill(white)
pygame.mouse.set_visible(False)
x,y = pygame.mouse.get_pos()
x = x - cursorPng.get_width()/2
y = y - cursorPng.get_height()/2
screen.blit(cursorPng,(x,y))
if x < 50 and y < 250:
screen.blit(enemy,(100,100))
for event in pygame.event.get():
...
答案 1 :(得分:-1)
你需要在屏幕上blit一个Surface和一个Rect。
首先,使用我用于加载图片的代码段。它确保正确加载图像:
def loadImage(name, alpha=False):
"Loads given image"
try:
surface = pygame.image.load(name)
except pygame.error:
raise SystemExit('Could not load image "%s" %s' %
(name, pygame.get_error()))
if alpha:
corner = surface.get_at((0, 0))
surface.set_colorkey(corner, pygame.RLEACCEL)
return surface.convert_alpha()
其次,当你得到Surface时,得到它的矩形:
cursorSurf = loadImage('resources/images/cursor.png')
cursorRect = cursorSurf.get_rect()
然后,在更新内部执行以下操作:
cursorRect.center = pygame.mouse.get_pos()
最后,像这样盲目地屏幕:
screen.blit(cursorSurf, cursorRect)
现在您将注意到您的鼠标无需移动鼠标即可正确呈现。