当不再满足 if 语句时，如何阻止 onkeypress 寻找按键，即如何暂时停用 onkeypress我是新的

问题描述

我对所有字母都这样做了，整个序列都在一个函数中。在变量“字母”的第一次迭代期间，它会正常工作，但在第二次迭代、第三次等期间，代码将为变量字母的过去迭代运行 fix_fall 函数。例如，如果第一次 letter == "a" fix_fall 只会在您单击 "a" 时运行，但是当您将 letter 更改为等于 "f" 时，它会在您单击 "a" 或 "f" 时运行 fix_fall。如何让“a”的按键停止激活，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")

解决方法

我不知道这是否解决了问题，因为我不明白问题是什么（当您添加更多信息时，我很乐意编辑答案），但与此同时，我对这个程序进行了一些有趣的编码：它会在屏幕上的随机 x 位置写下您按下的字母（这有点像您发布的片段中想要执行的操作）。

请注意，我避免使用一个不寻常的构造（无论如何对于 python）复制代码 26 次：closure。闭包究竟是什么超出了本答案的范围，但您可以将其（从极其广泛的意义上说）视为绑定到变量的函数。

函数 closure_writer 创建一个新函数，绑定到您调用 letter 的值 closure_writer，然后返回 那个功能。

对于每个字母，函数 closure_writer 被调用，返回的函数被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！