Python开发Hangman游戏如何仅在选中指定单词时激活对应代码段

核心错误原因 #

• 所有和tree相关的判定逻辑没有加当前选中目标单词为tree的前置条件，不管随机抽到的是sun还是tree，只要输入tree都会触发正确判定
• 存在多处冗余重复的letters == "tree"判断，布尔值拼写错误（Python中布尔值为True不是true）、赋值符号误用（把赋值用的=写成了比较用的==）
• 错误输入判定逻辑写法错误，永远无法触发
• 初始的字母猜对提示逻辑写在用户输入循环外，不会生效

修复后完整代码 #

#hangman mini-project
import random
import string
import time

hangmanwords = ['tree','sun']
sunchoices = ['s _ _', '_ u _', '_ _ n']
treechoices = ['t _ _ _', '_ r _ _', ' _ _ e _', '_ _ _ e']
limbCount = 5

hangmanword = random.choice(hangmanwords)
if hangmanword == "sun":
    print(random.choice(sunchoices))
else:
    print(random.choice(treechoices))

while True:
    letters = input("Please enter a letter to guess the word: ")
    print(f"The letter that you chose was {letters}")
    
    # 目标单词是tree时才执行tree相关逻辑
    if hangmanword == "tree":
        # 直接猜中完整单词
        if letters == "tree":
            input("Correct! The word was tree! Press enter to play again.")
            time.sleep(1)
            break
        # 猜中单个字母
        if letters == "r":
            print("Nice! You guessed one letter from the word!\n t r _ _")
        elif letters == "e":
            print("Wow! You guessed two letters from the word!\n t _ e e")
        # 猜错字母
        elif letters not in ['t','r','e']:
            print("Sorry, that's not a correct letter, feel free to try again.")
            limbCount -=1
    
    # 目标单词是sun时才执行sun相关逻辑
    elif hangmanword == "sun":
        # 直接猜中完整单词
        if letters == "sun":
            input("Correct! The word was sun! Press enter to play again.")
            time.sleep(1)
            break
        # 猜中单个字母
        if letters == "s":
            print("Nice! You guessed one letter from the word!\n s _ _")
        elif letters == "u":
            print("Nice! You guessed one letter from the word!\n _ u _")
        elif letters == "n":
            print("Nice! You guessed one letter from the word!\n _ _ n")
        # 猜错字母
        elif letters not in ['s','u','n']:
            print("Sorry, that's not a correct letter, feel free to try again.")
            limbCount -=1
    
    # 耗尽尝试次数
    if limbCount == 0: 
        print("Unfortunately, you are out of tries, better luck next time!")
        time.sleep(1)
        exit()

关键修改说明 #

• 给tree、sun对应的处理逻辑分别增加了目标单词匹配的前置判断，目标单词不对时对应逻辑完全不会执行，解决了输入tree在sun局也判定正确的问题
• 删掉了冗余重复的判断代码，修正了布尔值、赋值符号的写法错误
• 调整了错误提示的判定逻辑，现在可以正确识别猜错的字母
• 补充了sun局的猜词逻辑，游戏可以正常运行两个单词的猜词流程

内容的提问来源于stack exchange，提问作者Infui0ayaan