Python可搜索元组添加yes/no循环时变量未定义问题求解

• 字符串小写转换方法调用错误：获取用户输入后调用lower时未添加括号，导致yesno存储的是lower方法本身的引用而非转换后的小写字符串，所有条件判断都会失败进入else分支，部分运行环境在递归调用时会触发作用域异常，报yesno未定义错误。
• 变量名不规范：使用Python内置类型名tuple作为自定义变量名，会覆盖内置的元组类型，可能引发后续不可预期的错误。
• 冗余类型转换：input()本身返回值就是字符串类型，不需要额外用str()包裹。

# 替换为实际需要搜索的元组内容
search_tuple = ("apple", "banana", "orange", "grape")

for i in search_tuple:
    print(i, end=' ')
print()

def main():
    enter_name = input('Enter Search:')
    if enter_name in search_tuple:
        print('Yes its in there')
    else:
        print('No  its not there')

main()

def yes_or_no():
    # 给lower方法添加括号，正确执行小写转换
    yesno = input("Would You Like to Search Again?").lower()
    yesChoice = ['yes', 'y']
    noChoice = ['no', 'n']
    if yesno in yesChoice:
        main()
        yes_or_no()
    elif yesno in noChoice:
        print("Goodbye")
    else:
        print("Please Enter either yes or no")
        yes_or_no()
    
yes_or_no()

原代码用递归实现重复询问逻辑，递归深度过高时会触发栈溢出错误，建议改用while循环实现更稳定：

search_tuple = ("apple", "banana", "orange", "grape")
for i in search_tuple:
    print(i, end=' ')
print()

while True:
    enter_name = input('Enter Search:')
    if enter_name in search_tuple:
        print('Yes its in there')
    else:
        print('No  its not there')
    
    # 处理是否继续搜索的逻辑
    while True:
        yesno = input("Would You Like to Search Again?").lower()
        if yesno in ['yes', 'y']:
            break
        elif yesno in ['no', 'n']:
            print("Goodbye")
            exit()
        else:
            print("Please Enter either yes or no")

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