在Pygame中以光標方向繪制無限長的線

Question

我正在尋找有關pygame的幫助。 我正在用Python開發簡單的游戲來學習Pygame。 我想做一個可以旋轉並可以用激光線射擊的太空飛船。 我已經完成了箭頭鍵的控制，我們也可以用鼠標的位置旋轉飛船，但是拍攝時遇到了問題。 我想用從飛船位置到鼠標方向的無限長的線。 我該怎么做？ 這是我的代碼：

  def draw_objects(self):
        SCREEN.fill(BLACK)
        self.target = pygame.mouse.get_pos()
        self.x = self.player.position[0] #player x position
        self.y = self.player.position[1] #player y position
        self.mx = self.target[0] #mouse x position
        self.my = self.target[1] #mouse y position
        self.slope=float(float(self.y-self.my)/float(self.x-self.mx+0.1)) #slope
        self.x_new =  DISPLAY_WIDTH #ray length
        self.y_new = self.y + self.slope * (self.x_new - self.x) 
        self.player.draw()
        self.draw_columns()
        for agent in self.all_agents:
            agent.draw()
            agent.draw_vectors()
        if self.player.shoot == True:
            pygame.draw.line(SCREEN, GREEN, self.player.position,(self.x_new, self.y_new), 2)

        pygame.display.update()

它工作不正常，因為它只能在飛船的右邊工作。 在其他情況下，它會向光標反射出一條線。

感謝您的幫助。

Answer 1

slope不能保持方向。
您必須獲得player_x - mouse_x + 0.1符號並與x_new使用

    dx = player_x - mouse_x + 0.1

    reversed_sign_x = 1 if dx < 0 else -1

    x_new = reversed_sign_x * DISPLAY_WIDTH

---

完整的工作示例：

• 移動鼠標移動線，
• 單擊左按鈕設置玩家新位置。

。

import pygame

# --- constants ---

BLACK = (0, 0, 0)
GREEN = (0, 255, 0)
DISPLAY_WIDTH = 800
DISPLAY_HEIGHT = 600

# --- functions ---

def calculate(player_x, player_y, mouse_x, mouse_y):
    dx = player_x - mouse_x + 0.1
    dy = player_y - mouse_y

    reversed_sign_x = 1 if dx < 0 else -1
    #reversed_sign_y = 1 if dy < 0 else -1

    slope = dy/dx

    x_new = reversed_sign_x * DISPLAY_WIDTH
    y_new = player_y + slope * (x_new - player_x)

    return x_new, y_new

# --- main ---

# - init -

pygame.init()
SCREEN = pygame.display.set_mode((DISPLAY_WIDTH, DISPLAY_HEIGHT))

# - objects -

player_x = DISPLAY_WIDTH // 2
player_y = DISPLAY_HEIGHT // 2

mouse_x = player_x
mouse_y = player_y

x_new, y_new = calculate(player_x, player_y, mouse_x, mouse_y)

# - mainloop -

clock = pygame.time.Clock()
running = True

while running:

    # - events -

    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            running = False
        elif event.type == pygame.MOUSEBUTTONDOWN:
            player_x, player_y = event.pos
        elif event.type == pygame.MOUSEMOTION:
            x_new, y_new = calculate(player_x, player_y, event.pos[0], event.pos[1])

    # - updates -

    # empty

    # - draws -

    SCREEN.fill(BLACK)
    pygame.draw.line(SCREEN, GREEN, (player_x, player_y), (x_new, y_new), 2)
    pygame.display.flip()

    # - FPS -

    clock.tick(25)

# - end -

pygame.quit()

Answer 2

furas在右邊，您必須檢查鼠標是在播放器的左側還是右側，如果DISPLAY_WIDTH在左側，則取反。 我來到了一個類似的解決方案：

def target_coords(position):
    x, y = position  # Unpack player position into x, y.
    mx, my = pygame.mouse.get_pos()  # Unpack mouse pos into mx, my.
    slope = (y-my) / (x-mx+0.1)
    # This is a conditional expression, similar to `if condition: ... `else: ... `.
    x_new = DISPLAY_WIDTH if x < mx else -DISPLAY_WIDTH
    y_new = y + slope * (x_new - x)
    return x_new, y_new

請注意，此函數僅計算目標坐標並返回它們（函數最好只做一件事）。 畫線以及其他功能。

還有另一種解決方案：您可以使用pygame向量，然后首先將向量計算到目標，對其進行規范化並按所需的長度（ DISPLAY_WIDTH ）進行縮放。

import pygame
from pygame.math import Vector2


pygame.init()
DISPLAY_WIDTH = 640
GREEN = pygame.Color('aquamarine1')
screen = pygame.display.set_mode((640, 480))
clock = pygame.time.Clock()
position = Vector2(300, 200)  # A pygame.math.Vector2 as the position.
done = False

while not done:
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            done = True

    screen.fill((30, 30, 30))
    pygame.draw.circle(screen, GREEN, (int(position.x), int(position.y)), 7)
    # Calculate the vector to the target by subtracting pos from mouse pos.
    # Normalizing it gives you a unit vector which you can scale
    # by multiplying it with the DISPLAY_WIDTH.
    target_vec = (pygame.mouse.get_pos()-position).normalize() * DISPLAY_WIDTH
    pygame.draw.line(screen, GREEN, position, position+target_vec, 2)

    pygame.display.flip()
    clock.tick(30)

pygame.quit()