我正在制作一个简单的太空游戏,并希望我的船(xwingImg)能够自由移动。当按下a,d和s键时,我需要旋转船。我认为这样可行:pg.transform.rotate(xwingImg,90),但似乎没有任何改变。
import sys
import random
import pygame as pg
pg.init()
screen = pg.display.set_mode((1280, 800))
display_width = 1280
display_height= 800
black= (0,0,0)
white= (255,255,255)
red = (255,0,0)
blue_violet = (138,43,226)
xwingImg = pg.image.load('X-Wing.bmp').convert()
tieImg= pg.image.load('tiefighter.png').convert()
space=pg.image.load('space.jpg').convert()
BG_image = pg.image.load('space.jpg').convert()
def main():
clock = pg.time.Clock()
# Surfaces/images have a `get_rect` method which
# returns a rect with the dimensions of the image.
player_rect = xwingImg.get_rect()
player_rect.center = ( 640,400 )
change_x = 0
change_y = 0
enemies = []
spawn_counter = 30
done = False
while not done:
for event in pg.event.get():
if event.type == pg.QUIT:
done = True
if event.type == pg.KEYDOWN:
if event.key == pg.K_d:
pg.transform.rotate(xwingImg, 90)
change_x = 5
if event.key == pg.K_w:
change_y = -5
if event.key == pg.K_s:
change_y= 5
if event.key == pg.K_a:
change_x = -5
if event.type == pg.KEYUP:
if event.key == pg.K_d and change_x > 0:
change_x = 0
if event.key == pg.K_a and change_x < 0:
change_x = 0
if event.key == pg.K_w and change_y<0:
change_y=0
if event.key == pg.K_s and change_y>0:
change_y=0
# Spawn enemies if counter <= 0 then reset it.
spawn_counter -= 1
if spawn_counter <= 0:
# Append an enemy rect. You can pass the position directly as an argument.
enemies.append(tieImg.get_rect(topleft=(random.randrange(1280), -800 )))
spawn_counter = 30
# Update player_rect and enemies.
player_rect.x += change_x
player_rect.y += change_y
for enemy_rect in enemies:
enemy_rect.y += 5
# Collision detection with pygame.Rect.colliderect.
if player_rect.colliderect(enemy_rect):
print('Collision!')
# Draw everything.
screen.blit(BG_image, (0,0))
for enemy_rect in enemies:
screen.blit(tieImg, enemy_rect)
screen.blit(xwingImg, player_rect)
if player_rect.x >display_width:
player_rect.x = 0
if player_rect.x < 0:
player_rect.x= 1280
if player_rect.y>display_height:
player_rect.y = 0
if player_rect.y < 0:
player_rect.y= 800
pg.display.flip()
clock.tick(40)
if __name__ == '__main__':
main()
pg.quit()
sys.exit()
答案 0 :(得分:1)
pygame.transform.rotate()返回一个带有旋转对象的新Surface。您没有使用返回的Surface。您不希望反复旋转图像,因为旋转需要填充返回的曲面以使其适合矩形。
旋转(表面,角度) - &gt;表面
未经过滤的逆时针旋转。 angle参数表示度,可以是任何浮点值。负角度量将顺时针旋转。
除非以90度为增量旋转,否则图像将被填充得更大以保持新尺寸。如果图像具有像素alpha,则填充区域将是透明的。否则,pygame将选择与Surface colorkey或topleft pixel值匹配的颜色。
我只是跟踪图像应该旋转的角度,然后当检测到旋转事件时改变该角度。然后在您的绘制步骤blit
中按该角度旋转图像。
player_image = pg.transform.rotate(xwingImg, angle)
player_rect = player_image.get_rect()
player_rect.center = player_pos
screen.blit(player_image, player_rect)