3 回答

TA贡献1859条经验 获得超6个赞
在 pygame 中,事件类型是枚举器常量。必须将此常量映射到相应的类。使用字典:
def Exit(event):
# [...]
def Keydown():
# [...]
eventhandler = {pygame.QUIT: Exit, pygame.KEYDOWN: Keydown}
当然,字典也可以动态生成甚至扩展。(例如:eventhandler[pygame.MOUSEBUTTONDOWN] = MouseDown)
必须将事件委托给在事件处理程序中分配的相应操作:
while True:
for event in pygame.event.get():
if event.type in eventhandler:
eventhandler[event.type](event)
例:
class EventHandler():
@staticmethod
def determine_event(event): pass
class Exit(EventHandler):
@staticmethod
def determine_event(event):
pygame.quit()
quit(0)
class Keydown(EventHandler):
@staticmethod
def determine_event(event):
print(event.key)
eventhandler = {pygame.QUIT: Exit, pygame.KEYDOWN: Keydown}
while True:
for event in pygame.event.get():
if event.type in eventhandler:
eventhandler[event.type].determine_event(event)
对于更通用的方法,可以在列表中管理处理程序。因此,可以将多个操作关联到一个事件类型:
def Exit(event):
# [...]
def Keydown1():
# [...]
def Keydown2():
# [...]
eventhandler = {pygame.QUIT: [Exit], pygame.KEYDOWN: [Keydown1, Keydown2]}
while True:
for event in pygame.event.get():
if event.type in eventhandler:
for target in eventhandler[event.type]:
target(event)
例:
class EventHandler():
targets = {}
@staticmethod
def add(type, event):
EventHandler.targets.setdefault(type, []).append(event)
@staticmethod
def notify(event):
if event.type in EventHandler.targets:
for target in EventHandler.targets[event.type]:
target.determine_event(event)
@staticmethod
def determine_event(event): pass
class Exit(EventHandler):
@staticmethod
def determine_event(event):
pygame.quit()
quit(0)
class KeydownPrint(EventHandler):
@staticmethod
def determine_event(event):
print(event.key)
class KeydownAction(EventHandler):
@staticmethod
def determine_event(event):
print("action")
EventHandler.add(pygame.QUIT, Exit)
EventHandler.add(pygame.KEYDOWN, KeydownPrint)
EventHandler.add(pygame.KEYDOWN, KeydownAction)
while True:
for event in pygame.event.get():
EventHandler.notify(event)
或者甚至是两个答案的组合(参见@AKX的答案):
class EventHandler:
targets = {}
def register(type):
def decorator(fn):
EventHandler.targets.setdefault(type, []).append(fn)
return decorator
def notify(event):
fnl = EventHandler.targets[event.type] if event.type in EventHandler.targets else []
for fn in fnl:
fn(event)
@EventHandler.register(pygame.QUIT)
def onExit(event):
pygame.quit()
quit(0)
@EventHandler.register(pygame.KEYDOWN)
def keydownPrint(event):
print(event.key)
@EventHandler.register(pygame.KEYDOWN)
def keydownAction(event):
print("action")
while True:
for event in pygame.event.get():
EventHandler.notify(event)

TA贡献1803条经验 获得超6个赞
你不需要使用带有静态方法的类,并且基于Rabbid76的答案进行构建,一种(恕我直言)优雅的方法是使用装饰器在dict中注册每个事件处理程序:
import pygame
event_handler_registry = {}
def register_event_handler(event):
def decorator(fn):
event_handler_registry[event] = fn
return fn
return decorator
@register_event_handler(pygame.QUIT)
def on_exit(event):
pygame.quit()
quit(0)
@register_event_handler(pygame.KEYDOWN)
def on_key_down(event):
print(event.key)
while True:
for event in pygame.event.get():
event_handler = event_handler_registry.get(event.type)
if event_handler:
event_handler(event)
else:
print(f"No event handler for {event}")

TA贡献1824条经验 获得超6个赞
也许你可以用一个字符串使用eval():
EVENTS = {
pygame.QUIT : "quit_handler()" ,
pygame.MOUSEBUTTONDOWN: "click_handler(event)"
}
for event in pygame.event.get:
eval (EVENTS.get(event.type))
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