我如何将输入文本框的结果分配给一个变量，然后传递给后端的一个函数？

Question

我在使用 tkinter 编写刽子手游戏时遇到了一些问题，前端和后端是分开的，并且都使用 OOP（我知道在后端使用 OOP 并不是真正必要的，它可能会'作为一个图书馆已经更好了，但我把它作为一个学习练习来做）。我是 tkinter 的新手，我想知道如何将输入文本框的结果分配给一个变量，然后传递给后端的一个函数。我还想知道我的 self.letter_row 和 self.guesses_left 标签是否会在后端更新时更新，尽管我已经在 game_gui() 开始时将它们分配到前端，我不确定他们是否会自动执行此操作。这是所有的前端代码...

from tkinter import *
from tkinter.ttk import *
import hangmanSkeleton as hm

class game_gui():

    def __init__(self,master):
        self.master = master
        master.title("Hangman")

        this_game = hm.Hangman()

        self.word = this_game.word
        self.letter_row = this_game.letter_row
        self.letters_guessed = this_game.letters_guessed
        self.guesses_left = this_game.guesses_left
        self.letter = StringVar()

        self.lbl_word = Label(master, text = "Welcome to Hangman!")
        self.lbl_word.grid(row = 0,padx = 10 , pady = 15, columnspan = 2)

        self.lbl_row = Label(master, text = self.letter_row)
        self.lbl_row.grid(row = 1, column = 0, padx = 15, pady = 10)

        self.lbl_guesses = Label(master, text = "Guesses Left: " + str(self.guesses_left))
        self.lbl_guesses.grid(row = 1, column = 1, padx = 15, pady = 10)

        self.entry_letter= Entry(master, textvariable = self.letter)
        self.entry_letter.grid(row = 2, padx = 10, pady = 20, columnspan = 2)

        self.guess_button = Button(master, text = "Guess Letter", command=lambda: this_game.guess_letter(self.letter))
        self.guess_button.grid(row = 3, padx = 10, pady = 10, columnspan = 2) 


root = Tk()
gui = game_gui(root)
root.mainloop()

这只是我试图在上面的 self.guess_button() 中开始工作的后端函数..

def guess_letter(self, letter): 
    try: 
        if type(letter) != str or len(letter) != 1 or letter not in ascii_lowercase:
            raise TypeError
        if letter in self.letters_guessed:
            raise ValueError
        elif letter in self.word:
            for count in range (len(self.word)):
                if letter==self.word[count]:
                    self.letter_row = self.letter_row[0:count] + letter + self.letter_row[count+1:]
        else:
            pass
        self.guesses_left = self.guesses_left-1 #take one away from guesses
    except TypeError:
        print("Value given is not a letter")
    except ValueError:
        print("Letter has already been guessed")

谢谢！

Answer 1

Comment: I still can't get my labels self.letter_row and self.guesses_left to update though

是的，这些变量未绑定到 Label，我建议改用 Label.configure(text=...。
注意：我只显示.this_game.letter_row!

添加函数 update_labels(... 和 .configure(... 标签文本：

def update_labels(self):
    self.lbl_row.configure(text=self.this_game.letter_row)
    # and so on

在 last 在 .__init__(... and .guess_letter(....
中调用此函数根本不需要变量 self.letter_row。

Question: how i can assign the result of the entry textbox to a variable to then pass into a function in the backend.

更改以下内容：

使您的变量 this_game 成为 class game_gui
的成员
```
self.this_game = hm.Hangman()
```
将辅助方法 guess_letter(... 添加到 class game_gui。
把信传给.this_game.guess_letter(....
清除 .letter.
中的字母
```
def guess_letter(self, letter):
    self.this_game.guess_letter(letter.get())
    letter.set('')
```

更改command，指向辅助方法：

..., command=lambda: self.guess_letter(self.letter)

我如何将输入文本框的结果分配给一个变量，然后传递给后端的一个函数？

how i can assign the result of the entry textbox to a variable to then pass into a function in the backend?

python

oop

tkinter

widget

tkinter-entry