Question

所以我的程序是按行显示按钮，然后等待用户点击一个，但是会发生什么，它会返回主菜单。

我不确定如何使程序等待（我尝试了time.sleep（）和os.system（＆＃34;暂停＆＃34;），但是这两个都只是等待一定的时间和然后还返回主菜单 - 所以我使用控制台输入，只是为了看看盒子是否正确显示。

然而，现在使用控制台输入，单击任何按钮会使程序无响应（当按钮应该改变颜色时） - 我已经包含了一个屏幕截图。 Program not responding

我想知道自己做错了什么，我知道这有点乱，我对某种回应感到非常高兴，如果需要，我可以提供更多信息。非常感谢。

这是显示按钮的代码。

screen.fill(white) #clear the screen
xPos = -135;yPos = 200;tasktodo = "null" #set the position of where the first buttons should be placed
for i in tasklist[posInList][2]: #For each task the user has been set
    for y in QuestionTypes: #For each item in QuestionTypes ([MPQ,DBQ,RatioQ]...)
        if str(y) in str(i): #If the any item in QuestionTypes is in the tasks the user is set:
            task = (i.split(",")[0]) #This splits up the various tasks to the form of [DBQ,MPQ,AFQ]

            if xPos > display_height: #If the place a button would be placed is out of bounds
                xPos = 50 #Reset xPos
                yPos += yPos -82 #Lower the position of the button to another row
            else:
                xPos += 185 #Else, place the button 185 pixels to the right
            button(str(task), xPos, yPos, 150, 90, grey, lightGrey, action=str(task)) #Place a button with the name of the task, in position (xPos,yPos), dimensions (150,90),active/inactive colors of grey/light grey action being the task

            listOfSetQuestions.append(y) #Append the tasks the user has to do to a list
            amountoftasks += 1 #Keep track of the amount of tasks the user has to do.
# After the tasks are all printed, the user must input what task they wish to do
message_to_screen("These are the tasks your teacher has set, pick one:", black, "medium", -200) #send a message to the screen, black color, medium size font and y position of -200
pygame.display.update()

x = input("press ENTER to continue")

这是按钮的代码，因此您可以查看更多上下文。

def button(text, x, y, width, height, inactiveColor, activeColor, action = None):
cur = pygame.mouse.get_pos() #Gets the position of the cursor
click = pygame.mouse.get_pressed() #will return a tuple of 3, click[0] is left click, click[1] is middle and click[2] is right.
if x + width > cur[0] > x and y + height > cur[1] > y: #If the mouse is clicked and is between the borders of the button.
    pygame.draw.rect(screen, activeColor, (x,y,width,height)) #Change the color of the button to the active color, to show it is clicked
    if click[0] == 1 and action != None: #if the mouse is clicked, and there is a purpose to the click (clicked on a button)
            if action == "Create":
                Create()
            if action == "Sign in":
                Sign_In()
            if action == "Teacher":
                pass
            if action in QuestionTypes:
                print (action)
                global tasktodo
                tasktodo = action

else:
    pygame.draw.rect(screen, inactiveColor, (x,y,width,height))

text_to_button(text,black,x,y,width,height)

Answer 1

当要求input时，程序停止并等到用户输入内容。当程序停止时，它无法处理使你的操作系统相信程序崩溃的事件。因此，当使用pygame时，您无法获得控制台输入（除非您创建进程）。

如果你需要的只是延迟，你可以使用while - 循环

def wait(time):
    clock = pygame.time.Clock()

    while time > 0:
        dt = clock.tick(30) / 1000  # Takes the time between each loop and convert to seconds.
        time -= dt
        pygame.event.pump()  # Let's pygame handle internal actions so the os don't think it has crashed.

如果您希望用户决定等待多长时间，您可以检查用户事件

def wait_user_response():
    clock = pygame.time.Clock()
    waiting = True

    while waiting:
        clock.tick(30)

        for event in pygame.event.get():
            if event.type == pygame.KEYDOWN:  # Or whatever event you're waiting for.    
                waiting = False

当然还有其他更好的方法，但由于我不知道你的程序如何运作，这就是我所能提供的。

Pygame程序停止响应，没有错误

1 个答案: