程序不会根据输入调用正确的函数。不知道我错过了什么

Question

编辑：在 main() 中玩了一会儿之后，似乎程序正在选择 if/elif 块中调用的函数，而不管输入如何。我不知道为什么它在工作正常后开始这样做

我一遍又一遍地检查我的代码，但我无法弄清楚我没有看到什么。这是一个数字序列游戏，它要求用户在简单和困难之间选择一个难度。它在某一时刻工作得很好，但现在无论选择什么，它每次都会进入简单模式。即使您根本没有任何输入就按下回车键。

#main program function and difficulty selection
def main():
print('-----------------------------------------------')
print('Please choose a difficulty.')
difficulty = str(input('(e)asy|(h)ard: '))
print('-----------------------------------------------')
if difficulty == 'easy'or'e':
     easy()
elif difficulty == 'hard'or'h':
     hard()

然后我有一个简单的函数和一个困难的函数。 Hard 只是 easy 函数，只改变生成的序列的大小，没有别的。我检查了每个块，没有任何会影响调用哪个函数的更改。

无论游戏玩了多少次都会发生，所以一定是我的 main() 函数有问题

如果有帮助的话，剩下的代码就在这里，也许我遗漏了一些明显的东西。

import random
def easy():
    print ('Easy Mode','\n')
  #Generates inital number, step value, and upper limit
    num_sequence = 5
    numbers = random.randint(1,101)
    step = random.randint(1,20)
  #Accumulates and prints all but last number in sequence
    for num_generated in range (1, num_sequence):
        print(numbers)
        numbers = numbers + step
  #Sets tries allowed and subtracts wrong attempts
    guesses = 3
    while guesses > 0:
        user_num = int(input('Next Number: '))
        guesses = guesses - 1
        if user_num != numbers:
            if guesses == 0:
                break
            else:
                print('Try Again (Attempts Remaining:', guesses,')')              
        if user_num == numbers:
            break

#Prints appropriate message based on game results                      
  if user_num == numbers:
      print ('Correct','\n')
  if user_num != numbers:
       print ('Attempts Exceeded: The answer was',numbers,'\n')

#block for hard difficulty (same as above, sequence size changed to 4)
def hard():
    print ('Hard Mode','\n')
    num_sequence = 4

#main program function and difficulty selection
def main():
    print('-----------------------------------------------')
    print('Please choose a difficulty.')
    difficulty = str(input('(e)asy|(h)ard: '))
    print('-----------------------------------------------')
    if difficulty == 'easy'or'e':
         easy()
    elif difficulty == 'hard'or'h':
         hard()

#block for replay selection
replay = 'y'
while replay == 'y':
    main()
    replay = input('Play again? (y)|(n): ',)
    print ('\n')
    if replay == 'n':
        print('-----------------------------------------------')
        print('Goodbye!')
        print('-----------------------------------------------')
        break

hard() 与前几个

之后逐行的 easy() 代码相同

Answer 1

当您进行复合比较（使用 or）时，条件的两边都必须完整。也就是说，

if difficulty == 'easy'or difficulty == 'e':
    easy()
elif difficulty == 'hard'or difficulty == 'h':
    hard()

否则你会说 "if difficulty == 'easy' >> which is false, then assign difficulty to 'e'" 这不是本意。

程序不会根据输入调用正确的函数。不知道我错过了什么

Program won't call correct function based on input. No idea what I'm missing

module

input

function

call