Python 如何在pygame中实现两个输入框,而不必重复代码
上面是一个工作函数,用于制作一个屏幕,屏幕上有一个文本框。现在,您如何在同一屏幕上创建两个文本框,而不只是复制和粘贴同一代码两次 我的想法是,点击或激活的文本框将完成事件部分的内容,因此您不必重复两次。然而,我不知道如何实现这一点Python 如何在pygame中实现两个输入框,而不必重复代码,python,textbox,pygame,Python,Textbox,Pygame,上面是一个工作函数,用于制作一个屏幕,屏幕上有一个文本框。现在,您如何在同一屏幕上创建两个文本框,而不只是复制和粘贴同一代码两次 我的想法是,点击或激活的文本框将完成事件部分的内容,因此您不必重复两次。然而,我不知道如何实现这一点 提前感谢您可以创建一个类,并根据需要多次实例化不同的输入框 大概是这样的: def InputBox(): font = pygame.font.Font(None, 32) inputBox = pygame.Rect(50, 50, 140, 3
提前感谢您可以创建一个类,并根据需要多次实例化不同的输入框 大概是这样的:
def InputBox():
font = pygame.font.Font(None, 32)
inputBox = pygame.Rect(50, 50, 140, 32)
colourInactive = pygame.Color('lightskyblue3')
colourActive = pygame.Color('dodgerblue2')
colour = colourInactive
text = ''
active = False
isBlue = True
while True:
for event in pg.event.get():
if event.type == pg.QUIT:
pg.quit()
sys.exit()
if event.type == pygame.MOUSEBUTTONDOWN:
if inputBox.collidepoint(event.pos):
active = not active
else:
active = False
colour = colourActive if active else colourInactive
if event.type == pygame.KEYDOWN:
if active:
if event.key == pygame.K_RETURN:
print(text)
text = ''
elif event.key == pygame.K_BACKSPACE:
text = text[:-1]
else:
text += event.unicode
screen.fill(screenGray)
txtSurface = font.render(text, True, colour)
width = max(200, txtSurface.get_width()+10)
inputBox.w = width
screen.blit(txtSurface, (inputBox.x+5, inputBox.y+5))
pygame.draw.rect(screen, colour, inputBox, 2)
if isBlue:
color = (0, 128, 255)
else:
color = (255, 100, 0)
pg.display.flip()
clock.tick(60)
InputBox()
#伪代码#
类输入框(pygame.Rect):
定义初始化(self,position=(50,50,140,32),
font=pygame.font.font(无,32),
color\u inactive=pygame.color('lightskyblue3'),
color\u active=pygame.color('dodgerblue2'),
文本=“”,
活动=错误)
super()
大概是这样的:
def InputBox():
font = pygame.font.Font(None, 32)
inputBox = pygame.Rect(50, 50, 140, 32)
colourInactive = pygame.Color('lightskyblue3')
colourActive = pygame.Color('dodgerblue2')
colour = colourInactive
text = ''
active = False
isBlue = True
while True:
for event in pg.event.get():
if event.type == pg.QUIT:
pg.quit()
sys.exit()
if event.type == pygame.MOUSEBUTTONDOWN:
if inputBox.collidepoint(event.pos):
active = not active
else:
active = False
colour = colourActive if active else colourInactive
if event.type == pygame.KEYDOWN:
if active:
if event.key == pygame.K_RETURN:
print(text)
text = ''
elif event.key == pygame.K_BACKSPACE:
text = text[:-1]
else:
text += event.unicode
screen.fill(screenGray)
txtSurface = font.render(text, True, colour)
width = max(200, txtSurface.get_width()+10)
inputBox.w = width
screen.blit(txtSurface, (inputBox.x+5, inputBox.y+5))
pygame.draw.rect(screen, colour, inputBox, 2)
if isBlue:
color = (0, 128, 255)
else:
color = (255, 100, 0)
pg.display.flip()
clock.tick(60)
InputBox()
#伪代码#
类输入框(pygame.Rect):
定义初始化(self,position=(50,50,140,32),
font=pygame.font.font(无,32),
color\u inactive=pygame.color('lightskyblue3'),
color\u active=pygame.color('dodgerblue2'),
文本=“”,
活动=错误)
super()。然后你可以多次使用它
完整的工作示例
# pseudocode #
class InputBox(pygame.Rect):
def __init__(self, position=(50, 50, 140, 32),
font=pygame.font.Font(None, 32),
color_inactive=pygame.Color('lightskyblue3'),
color_active=pygame.Color('dodgerblue2'),
text = '',
active=False)
super().__init__(position) #<- this may need to be adjusted to pygame Rect specs
self.font = font
self.color_inactive = color_inactive
self.color_active = color_active
self.color = color_inactive
self.text = text
self.active = active
顺便说一句:对于更多输入,您可以将它们保留在列表中(如GUI框架中)
您可以使用绘制类和处理事件的方法创建类。然后你可以多次使用它
完整的工作示例
# pseudocode #
class InputBox(pygame.Rect):
def __init__(self, position=(50, 50, 140, 32),
font=pygame.font.Font(None, 32),
color_inactive=pygame.Color('lightskyblue3'),
color_active=pygame.Color('dodgerblue2'),
text = '',
active=False)
super().__init__(position) #<- this may need to be adjusted to pygame Rect specs
self.font = font
self.color_inactive = color_inactive
self.color_active = color_active
self.color = color_inactive
self.text = text
self.active = active
顺便说一句:对于更多输入,您可以将它们保留在列表中(如GUI框架中)