当onkeypress的if语句不再满足时，我如何停止onkeypress寻找keypress，例如，如何临时停用onkeypress (im new)

Question

我对所有的字母都这样做了，整个序列都在一个函数中。在变量" letter“的第一次迭代期间，它将正常工作，但在第二次迭代、第三次迭代等期间，代码将为变量letter的过去迭代运行fix_fall函数。例如，如果第一次使用letter == "a“，fix_fall将只在您单击"a”时运行，但当您将letter更改为等于"f“时，它将在您单击"a”或"f“时运行fix_fall。我如何让"a“的onkeypress停止激活，直到字母再次等于"a”。

lenny = 0
def draw_an_letter():
  global letter
  global lenny
  apple_xcor = apple.xcor()
  apple_ycor = apple.ycor()
  drawer_xcor = apple_xcor - 14
  drawer_ycor = apple_ycor - 32
  if lenny > 0:
    del letter
    letter = random.choice(letters)
    letter = letter.lower()
  drawer.goto(drawer_xcor, drawer_ycor)
  drawer.color("white")
  drawer.write(letter, font=("Arial", 35, "bold"))
  lenny = lenny + 1 

def fix_fall():
   fall_apple(apple)
   apple_newxcor = random.randint(-130,150)
   apple.goto(apple_newxcor, 40)
   apple.showturtle()
   draw_an_letter()
   wn.listen()
   if letter == "a":
      wn.onkeypress(fix_fall, "a")
   if letter == "b":
      wn.onkeypress(fix_fall, "b")
   if letter == "c":
      wn.onkeypress(fix_fall, "c")
   if letter == "d":
      wn.onkeypress(fix_fall, "d")
   if letter == "e":
      wn.onkeypress(fix_fall, "e")
   if letter == "f":
      wn.onkeypress(fix_fall, "f")
   if letter == "g":
      wn.onkeypress(fix_fall, "g")
   if letter == "h":
      wn.onkeypress(fix_fall, "h")
   if letter == "i":
      wn.onkeypress(fix_fall, "i")
   if letter == "j":
      wn.onkeypress(fix_fall, "j")
   if letter == "k":
      wn.onkeypress(fix_fall, "k")
   if letter == "l":
      wn.onkeypress(fix_fall, "l")
   if letter == "m":
      wn.onkeypress(fix_fall, "m")
   if letter == "n":
      wn.onkeypress(fix_fall, "n")
   if letter == "o":
      wn.onkeypress(fix_fall, "o")
   if letter == "p":
      wn.onkeypress(fix_fall, "p")
   if letter == "q":
      wn.onkeypress(fix_fall, "q")
   if letter == "r":
      wn.onkeypress(fix_fall, "r")
   if letter == "s":
      wn.onkeypress(fix_fall, "s")
   if letter == "t":
      wn.onkeypress(fix_fall, "t")
   if letter == "u":
      wn.onkeypress(fix_fall, "u")
   if letter == "v":
      wn.onkeypress(fix_fall, "v")
   if letter == "w":
      wn.onkeypress(fix_fall, "w")
   if letter == "x":
      wn.onkeypress(fix_fall, "x")
   if letter == "y":
      wn.onkeypress(fix_fall, "y")
   if letter == "z":
      wn.onkeypress(fix_fall, "z")

Answer 1

我不知道这是否解决了问题，因为我不知道问题是什么(当你添加更多信息时，我会很高兴地编辑答案)，但同时，我在编写这个程序时获得了一些乐趣:它会在屏幕上随机的x位置写下你按下的字母(这看起来有点像你想从你发布的代码片段中做的事情)。

请注意，我使用了一个不同寻常的构造(至少对于python )：一个closure，从而避免了重复26次代码。闭包的确切含义超出了这个答案的范围，但您可以将其(在非常宽泛的术语中)看作是绑定到变量的函数。

function closure_writer 将创建一个新函数，并将其绑定到您调用closure_writer的值letter，然后返回该函数。

对于每个字母，都会调用函数closure_writer，并将返回的函数链接为该字母的回调。

import random
from string import ascii_lowercase
import turtle


def random_xpos():
    """Move the turtle to a random x position."""
    newxcor = random.randint(-600, 600)
    t.setx(newxcor)


def write_letter(letter):
    """Write a letter in the current spot."""
    t.write(letter, font=("Arial", 35, "bold"))


def closure_writer(letter):
    """Return a function that writes the specified letter."""

    def writer():
        """Write the specified letter."""
        random_xpos()
        write_letter(letter)

    # note that we are NOT calling the function, but returning it
    return writer


# setup the turtle
t = turtle.Turtle()
t.penup()
turtle.listen()

# register all the keypress
for letter in ascii_lowercase:
    # closure_writer returns a new function that will be called at each keypress
    # and each function is created with its own letter associated
    turtle.onkeypress(closure_writer(letter), letter)

# start the loop
turtle.mainloop()

干杯!