让Tkinter等到按下按钮

问题描述 投票:3回答:2

我有一个游戏,当一个按钮被创建时,我需要我的程序只显示这个屏幕,直到他们按下“下一级”所有这些代码都在一个while循环中,所以在一个大的while循环控制游戏。

......

if game.playerDistance >= game.lowerBound() and game.playerDistance <= game.upperBound():
        game.level += 1

        showLevelResults(game)

        #NextLevelButton
        btnNextLevel = Button(root,
                    #Random Config
                    command = nextLevel,
                    )
        btnNextLevel.place(x=1003, y=492, anchor=NW,  width=247, height=78)

        updateMainScreen()

        while nextLev == False:
            #What Do I put in here to force a wait
    else:

......

nextLev = False
def nextLevel():
    nextLev = True

...

目前这使它保持在while循环中并且当按下按钮时没有任何变化我使用time.sleep(1)来保持等待并且还打印等待btn按下,但是这会使控制台发出警告并且当按下按钮时仍然没有改变屏幕。

def showGameSurvival():

game = gamemode_normal()

while game.health != 0:
    game.next = False 
    clearScreen()
    changeBackground("Survival")

    #Placing Labels on the screen for game.....

    #... Health
    root.update()

    lblCountDownLeft = Label(root, bg="White", fg="Green", font=XXLARGE_BUTTON_FONT)
    lblCountDownLeft.place(x=169, y=350, anchor=CENTER)
    lblCountDownRight = Label(root, bg="White", fg="Green", font=XXLARGE_BUTTON_FONT)
    lblCountDownRight.place(x=1111, y=350, anchor=CENTER)
    #CountDown
    count = 7
    while count > 0:                
        lblCountDownLeft['text'] = count
        lblCountDownRight['text'] = count
        root.update()
        count -= 1
        time.sleep(1)

    lblCountDownLeft.destroy()
    lblCountDownRight.destroy()
    root.update()
    #Num on left x=169, right, x=1111 y=360

    game.measureDistance()
    if game.playerDistance >= game.lowerBound() and game.playerDistance <= game.upperBound():
        game.level += 1
        clearScreen()
        changeBackground("Survival")
        graphicalDisplay(game)

        #NextLevelButton
        btnNextLevel = Button(root,
                    bg= lbBlue,
                    fg="white",
                    text="Level" + str(game.level),
                    font=SMALL_BUTTON_FONT,
                    activebackground="white",
                    activeforeground= lbBlue,
                    command= lambda: nextLevel(game),
                    bd=0)
        btnNextLevel.place(x=1003, y=492, anchor=NW,  width=247, height=78)
        root.update()
        while game.next == False:
            print(game.next)
    else:
        game.health -= 1

    if game.allowance > 4:
        game.allowance = int(game.allowance*0.9)

#when game is over delete the shit        
if game.health == 0:
    del game

下一个按钮现在调用此函数:def nextLevel(game): game.next = True

python python-3.x user-interface button tkinter
2个回答
10
投票

让tkinter等待某个事件的最简单方法是调用其中一个“等待”函数,例如wait_variablewait_windowwait_visibility

在您的情况下,您希望等待按钮单击,因此您可以使用wait_variable然后让按钮设置变量。当您单击按钮时,将设置变量,并且当设置变量时,将返回对wait_variable的调用。

例如:

import tkinter as tk
root = tk.Tk()
...
var = tk.IntVar()
button = tk.Button(root, text="Click Me", command=lambda: var.set(1))
button.place(relx=.5, rely=.5, anchor="c")

print("waiting...")
button.wait_variable(var)
print("done waiting.")

注意:您不必使用IntVar - 任何特殊的Tkinter变量都可以。此外,无论您将其设置为什么,该方法都会等到它发生变化。


1
投票

只是想添加评论,但没有足够的声誉。我无法让wait_variable(var)在按钮上为我工作。我必须在我的框架上使用它。我猜这是Python 2和3之间的变化。

这是我得到的错误:

Traceback(最近一次调用最后一次):

文件“AutoPlotHydroData2.py”,第434行,在btnOK.wait_variable(okVar)中AttributeError:'NoneType'对象没有属性'wait_variable'

我的工作代码如下:

    # Launch frame to collect needed input values
    myFrame = tk.Tk()
    myFrame.configure(background='lightblue')

    # Add some widgets

    # OK Button
    okVar = tk.IntVar()
    btnOK = tk.Button(myFrame, text="Submit", pady=5, font=("Arial Bold", 10),
        bg='lightgray', command=lambda: okVar.set(1)).grid(row=14, column=0)

    # Show frame
    myFrame.tkraise()

    # Wait for OK button to be pressed
    #btnOK.wait_variable(okVar) - this didn't work
    myFrame.wait_variable(okVar)

    # Close frame
    myFrame.destroy()

我在循环中有这个代码来处理多个文件。希望这有助于某人。

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