我可以做些什么来修复我的简单电脑玩家游戏代码？语法错误

Question

我是 Python 新手，我想练习循环，因为我遇到了最大的麻烦。我决定制作一个游戏，让用户从 0-100 中选择一个数字，看看他们是否能战胜电脑。我现在要做的只是开始。代码未完成。但是在尝试代码时，我得到了一个语法错误，箭头指向elif函数上的冒号。我该如何解决？我能做些什么？我接受对我的代码的任何其他附加评论，以使其更好。这是我的代码：import randommin = 0max = 100roll_again = "yes"quit = "no"players_choice = input()computer = random.randintwhile roll_again == "yes":print("Pick a number between 1-100: ")print(players_choice)if players_choice >= 0:print("Your number of choice was: ")print(players_choice)print("Your number is high.")if computer >= 0:print("Computers number is: ")print(computer)print("Computers number is high.")if computer >= players_choice:print("Computer wins.")print("You lose.")print("Would you like to play again? ", +roll_again)elif: print(quit)end目标：修复电脑玩家游戏，同时了解更多关于 python 的信息。提供有关从何处开始的其他文档会很有帮助。

Answer 1

你得到一个错误指向的原因elif是因为elif需要一个条件来检查。你需要像这样使用if elif和else：

if a == b:

    print('A equals B!')

elif a == c:

    print('A equals C!')

else:

   print('A equals nothing...')

此外，Python 依靠缩进来确定什么属于什么，因此请确保您注意缩进（没有end）。

修复 if 语句和缩进后，您的代码有更多错误，但您应该能够查找帮助来修复这些错误。

Answer 2

有很多缩进问题，ifandelif语句使用不正确。还要看看 while 循环是如何工作的。

根据您在此处提供的代码，这是一个可行的解决方案，但还有许多其他方法可以实现这一点。

以下是一些关于if/else语句和其他初学者主题的有用教程：

Python IF...ELIF...ELSE 语句

import random

minval = 0

maxval = 100

roll_again = "yes"

quit_string = "no"

while True:

    players_choice = int(input("Pick a number between 1-100:\n"))

    computer = random.randint(minval,maxval)

    #checks if players choice is between 0 and 100

    if players_choice >= 0 and players_choice <= 100:

        print("Your number of choice was: ")

        print(players_choice)

    #otherwise it is out of range

    else:

        print("Number out of range")

    #check if computers random number is in range from 0 to 100        

    if computer >= 0 and computer <= 100:

        print("Computers number is: ")

        print(computer)

    # if computer's number is greater than players number, computer wins   

    if computer > players_choice:

        print("Computer wins.")

        print("You lose.")

    #otherwise if players number is higher than computers, you win

    elif computer < players_choice:

        print("You won.")

    #if neither condition is true, it is a tie game

    else:

        print("Tied game")

    #ask user if they want to continue        

    play_choice = input("Would you like to play again? Type yes or no\n")

    #checks text for yes or no use lower method to make sure if they type uppercase it matches string from roll_again or quit_string 

    if play_choice.lower() == roll_again:

        #restarts loop

        continue

    elif play_choice.lower() == quit_string:

        #breaks out of loop-game over

        break

Answer 3

你的代码有很多问题。这是一个工作版本，希望它可以帮助您理解一些概念。如果没有，请随时询问

import random

# min and max are names for functions in python. It is better to avoid using

# them for variables

min_value = 0

max_value = 100

# This will loop forever uless something 'breaks' the loop

while True:

  # input will print the question, wait for an anwer and put it in the

  # 'player' variable (as a string, not a number)

  player = input("Pick a number between 1-100: ")

  # convert input to a number so we can compare it to the computer's value

  player = int(player)

  # randint is a function (it needs parentheses to work)

  computer = random.randint(min_value, max_value)

  # print what player and computer chose

  print("Your choice: ", player)

  print("Computer's choice: ", computer)

  # display the results

  if computer >= player:

    print("Computer wins. You loose")

  else:

    print("you win.")

  # Determine if user wants to continue playing

  choice = raw_input("Would you like to play again? (yes/No) ")

  if choice != 'yes':

    # if not

    break