如何在pygame中旋转图像?

时间:2017-05-04 12:12:46

标签: python pygame

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()

1 个答案:

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)