Question

我只想让我的代码首先绑定到红色并使红色的处理程序在红色结束后重新绑定为黄色。林不知道如何做到这一点。有谁知道这是怎么做的。因为在我的代码中，如果您运行它。它只运行黄色函数而不是红色函数，因此我确定这是因为我覆盖了绑定函数。

from tkinter import *
import random
root = Tk()
root.geometry("300x300")
score = 0
lb1 = Listbox(root)
e = Entry(root)



def binder1():
    root.bind('<Return>', red)


def red(event):
    print(e.get())
    if e.get() == "red":
        lb1.insert("end", "Correct")
        binder1
        global score
        score += 1
        mb4['text'] = str("Score ") + str(score)


redcolour = Label(root, text = "What colour is this boi!", fg = "Red")


def yellow(event):
    print(e.get())
    if e.get() == "yellow":
        lb1.insert("end", "Correct")
        global score
        score += 1
        mb4['text'] = str("Score ") + str(score)

root.bind('<Return>', yellow)
yellowcolour = Label(root, text = "What colour is this boi!", fg = "Yellow")


mb4= Menubutton(root, text = str("Score: ") + str(score))
mb4.menu = Menu(mb4)





e.focus_set()
e.pack()
yellowcolour.pack()
redcolour.pack()
lb1.pack()
mb4.pack()
root.mainloop()

Answer 1

对于我来说不清楚，为什么你要先绑定红色。

我将使其更加动态，并让用户做出决定。

像这样：

from tkinter import *

def check_input(event):
    global score
    my_input = e.get()
    if my_input == "yellow":
        lb1.insert("end", "Correct yellow")
        score += 1
        mb4['text'] = str("Score ") + str(score)
    elif my_input == "red":
        lb1.insert("end", "Correct red")
        score += 1
        mb4['text'] = str("Score ") + str(score)
    else:
        return None


if __name__ == "__main__":
    root = Tk()
    root.geometry("300x300")
    root.bind('<Return>', check_input)
    score = 0

    lb1 = Listbox(root)
    e = Entry(root)
    redcolour = Label(root, text = "What colour is this boi!", fg = "Red")  
    yellowcolour = Label(root, text = "What colour is this boi!", fg = "Yellow")
    mb4= Menubutton(root, text = str("Score: ") + str(score))

    e.focus_set()

    e.pack()
    yellowcolour.pack()
    redcolour.pack()
    lb1.pack()
    mb4.pack()

    root.mainloop()

我如何在python中将一个函数绑定到另一个函数

1 个答案: