这是我正在尝试编写的一个简单的测试脚本,它将帮助我自学tkinter...
from tkinter import *
def hello():
print("U pressed it lol")
global window1, window2
window2 = None
window1 = None
def setWindow(windowEnter):
global window
window = windowEnter
window = Tk()
window.attributes("-fullscreen", True)
def newScreen(newScreen, screenToDelete):
setWindow(newScreen)
print("New Window Created")
screenToDelete.destroy()
print("He ded lol")
setWindow(window1)
def setStuff():
button = Button(window1, text="hey", command=hello)
label = Label(window1, text="YoYoYo My dude")
button2 = Button(window1, text="Next Page", command = lambda: newScreen(window2, window1))
button.pack()
label.pack()
button2.pack()
setStuff()
当我运行此代码时,它会返回错误?
File "C:Users 26341Desktoptest.py", line 19, in newScreen
screenToDelete.destroy()
AttributeError: 'NoneType' object has no attribute 'destroy'
为什么这不起作用以及我该如何解决它?
提前致谢:)(顺便说一句,我正在使用python 3.6(
你设置
window2 = None
window1 = None
作为全局变量,然后定义命令函数,button2
lambda: newScreen(window2, window1)
它调用newScreen
的值 window2 和 window1 都是None
的,因此错误。这里的根本问题是您的setWindow
函数:
def setWindow(windowEnter):
global window
window = windowEnter
window = Tk()
window.attributes("-fullscreen", True)
这不符合您使用它的方式。当你调用setWindow(window1)
时,你传递了window1
的值,函数对变量的作用在全局范围内是看不到的。一个简单的例子是这样的:
def increment(a):
a +=1
x = 1
print(x)
increment(x)
print(x)
这将打印 1 两次。
为了实现你想要的,我建议你使用字典来跟踪你的窗口。
from tkinter import *
def hello():
print("U pressed it lol")
global window1, window2
windows = {}
def setWindow(window_name):
windows[window_name] = Tk()
windows[window_name].attributes("-fullscreen", True)
def newScreen(newScreen_name, screenToDelete_name):
setWindow(newScreen_name)
print("New Window Created")
windows[screenToDelete_name].destroy()
del windows[screenToDelete_name] #delete invalid entry from dict
print("He ded lol")
setWindow("window1")
def setStuff():
button = Button(windows["window1"], text="hey", command=hello)
label = Label(windows["window1"], text="YoYoYo My dude")
button2 = Button(windows["window1"], text="Next Page", command = lambda: newScreen("window2", "window1"))
button.pack()
label.pack()
button2.pack()
setStuff()
注意:以前你的函数是def newScreen(newScreen, screenToDelete)
的,这是非常混乱/糟糕的风格,因为函数和它的第一个参数共享相同的名称。无论如何,我还是更改了它以强调它现在将字符串作为参数,但请记住它。
无法测试它,但我发现了一个错误来源:
lambda: newScreen(window2, window1)
这将创建一个不接受任何参数的lambda函数,因此window2和window1将是None,而None没有destroy((方法,因此错误。相反,请尝试:
lambda window2, window1: newScreen(window2, window1)