我正在建立一个照相平台，在这个目标中我的目标是进行以下过程： （显示说明并等待，直到OK区域上的鼠标按下） - ＆gt; RUN PHOTOBOOTH - ＆gt;显示图片：打印图片或拍摄另一张照片（超时xx秒返回主菜单）
我在显示图片方面有以下代码，我暂停代码以查看图像，或者图像显示但没有捕获鼠标点击
def PreviewMontage(MontageFile):
global LastTap
LastTap = time.time()
print("Session ID:", SessionID)
print("Show something.")
preview = pygame.image.load(MontageFile)
PILpreview = Image.open(MontageFile)
previewSize = PILpreview.size # returns (width, height) tuple
#added /1.5
ScaleW = AspectRatioCalc(previewSize[0]/1.5, previewSize[1]/1.5, SCREEN_HEIGHT)
preview = pygame.transform.scale(preview, (ScaleW, SCREEN_HEIGHT))
SetBlankScreen()
background.blit(preview, (SCREEN_WIDTH/2-ScaleW/2, 0))
PrintScreen()
#inserting conditions here - get mouse
camera.stop_preview()
UpdateDisplay()
Wait()
#sleep(20)
return
def PrintScreen():
#defines the text of the printscreen and buttons
#insert button for printing
pygame.draw.rect(background, rgbGREEN, pygame.Rect(NEXT_X, 0, ZONEWIDTH, SCREEN_HEIGHT))
#restarting button
pygame.draw.rect(background, rgbRED, pygame.Rect(PREV_X, PREV_Y, ZONEWIDTH, SCREEN_HEIGHT))
##text
Text = "Print or restart?"
Text = smallfont.render(Text, 1, rgbRED)
textpos = Text.get_rect()
textpos.centerx = background.get_rect().centerx
height = Text.get_height()
background.blit(Text,(textpos)) #Write the small text
return
# End of function.
我已经尝试过这段代码来等待输入（仅接受触摸屏左侧或右侧的输入） - 它正在工作但我似乎没有可靠的超时：
def Wait():
clock = pygame.time.Clock()
waiting = True
while waiting:
#clock.tick(time)
time = 60
#dt = clock.tick(30) / 1000 # Takes the time between each loop and convert to seconds.
#time -= dt
time = time -1
#print("Wainting..", waiting, "time ", time)
for event in pygame.event.get():
if event.type == pygame.MOUSEBUTTONDOWN and event.button == LEFTMOUSEBUTTON:
x, y = event.pos
print("You pressed the left mouse button at (%d, %d)" % event.pos)
LeftMouseButtonDown(x, y)
if time == 0:
waiting = False
return
Time变量不会减去...关于超时概念的任何想法？
这是＆＃34;主循环＆＃34;：
while running:
event = pygame.event.poll()
if event.type == pygame.QUIT:
running = 0
elif event.type == pygame.MOUSEBUTTONDOWN and event.button == LEFTMOUSEBUTTON:
x, y = event.pos
print("You pressed the left mouse button at (%d, %d)" % event.pos)
LeftMouseButtonDown(x, y)
elif event.type == pygame.KEYDOWN:
if event.key == pygame.K_F4:
print('F4 pressed, quitting.')
QuitGracefully()
这是鼠标检测（我无法在我的预览功能上工作）：
def LeftMouseButtonDown(xx, yy):
# Detect Taps in Previous Zone
if xx >= PREV_X and xx <= ZONEWIDTH:
TapPrev()
# Detect Taps in Next Zone
if xx >= NEXT_X and xx <= SCREEN_WIDTH:
TapNext()
# Detect Taps in the Start Zone
if xx >= START_MIN_X and yy >= START_MIN_Y and xx <= START_MAX_X and yy <= START_MAX_Y:
TapStart()
# Detect Taps in the Up Zone.
elif xx >= UP_MIN_X and yy >= UP_MIN_Y and xx <= UP_MAX_X and yy <= UP_MAX_Y:
KonamiCode('Up')
# Detect Taps in the Down Zone.
elif xx >= DOWN_MIN_X and yy >= DOWN_MIN_Y and xx <= DOWN_MAX_X and yy <= DOWN_MAX_Y:
KonamiCode('Down')
# Detect Taps in the Left Zone.
elif xx >= LEFT_MIN_X and yy >= LEFT_MIN_Y and xx <= LEFT_MAX_X and yy <= LEFT_MAX_Y:
KonamiCode('Left')
# Detect Taps in the Right Zone.
elif xx >= RIGHT_MIN_X and yy >= RIGHT_MIN_Y and xx <= RIGHT_MAX_X and yy <= RIGHT_MAX_Y:
KonamiCode('Right')
else:
KonamiCodeReset()
print("No Event")
return
# End of function.
答案 0 :(得分：0)
我认为你误解了事件是如何在pygame中处理的。
wait功能 wait方法给出错误'int' object is not iterable的原因是因为您试图循环整数。 event.type中的for event.pos in event.type:是一个整数，表示它是什么类型的事件（从主循环中读取event变量）。相反，您希望迭代（循环）自上次检查以来发生的所有事件。这可以通过调用事件队列来完成：for event in pygame.event.get():。
然后，您需要检查事件是否是您想要突破循环的事件类型。我不知道KonamiCode('Left')做了什么，但他们没有使用你迭代的事件，所以我怀疑他们会工作。相反，请检查if event.type == x其中x是您要退出循环的事件类型。
编辑：对更新后的功能的快速修复可能如下所示：
def wait(time):
"""
Makes the program halt for 'time' seconds or until the user press the left mouse button.
"""
clock = pygame.time.Clock()
waiting = True
while waiting:
dt = clock.tick(30) / 1000 # Takes the time between each loop and convert to seconds.
time -= dt
for event in pygame.event.get():
if event.type == pygame.MOUSEBUTTONDOWN and event.button == LEFTMOUSEBUTTON:
x, y = event.pos
print("You pressed the left mouse button at (%d, %d)" % event.pos)
LeftMouseButtonDown(x, y)
if time <= 0:
waiting = False
return # There is no need for explicit return
一些改进代码的提示。
event = pygame.event.poll()将仅返回事件队列中的单个事件。您可以改为调用pygame.event.get()，它将返回自上次调用以来的所有事件，然后遍历事件（如上所述和链接中所述。PreviewMontage加载图像，缩放图像，将其blit到背景，创建文本，blit文本，更新显示并等待30秒（或毫秒）。它有点多，使代码可重用性更低，可读性更低。lower_case等通用命名约定，为类使用CamelCase，可以帮助其他读者更轻松地识别名称，从而更容易理解和阅读代码。