如何执行不区分大小写的字符串匹配python 3.10?[复制]

How to perform case insensitive string-matching python 3.10? [duplicate]

提问人:Weverson Goncalves Muciarone 提问时间:10/6/2023 最后编辑:CausingUnderflowsEverywhereWeverson Goncalves Muciarone 更新时间:10/6/2023 访问量:52

问:

我正在尝试从 python 版本 3.10+ 运行一个 case 语句,用户应该在其中键入单词“add”作为输入。但是,如果用户使用大写字母(如“Add”),则不起作用。

todos = []

while True:
    user = input('Enter add, show, edit, complete or exit to finish. : ')
    user = user.strip() # This will change the string teh user typed and remove extra spaces

    match user:
        case 'add':
            todo = input('Add an item: ')
            todos.append(todo)
        case 'show' | 'display':
            for i, item in enumerate(todos):
                item = item.title()  # This will make all first letters capitalized
                print(f'{i + 1}-{item}\n')
                list_lengh = len(todos)
            print(f'You have {list_lengh} item in your to do list.\n')

我尝试在匹配用户中使用该方法,但没有奏效。我可以使用案例“添加”|“添加”,但我正在努力找到一个强大的解决方案。capitalize()

我希望用户能够从 case 语句中键入单词,例如“add”、“Add”甚至“ADD”。

python 字符串 匹配 不区分大小写

评论


答:

1赞 Andrej Kesely 10/6/2023 #1

使用户输入小写(带):str.lower

todos = []

while True:
    user = input("Enter add, show, edit, complete or exit to finish. : ")
    user = user.strip().lower()  # <-- put .lower() here

    match user:
        case "add":
            todo = input("Add an item: ")
            todos.append(todo)
        case "show" | "display":
            for i, item in enumerate(todos):
                item = item.title()  # This will make all first letters capitalized
                print(f"{i + 1}-{item}\n")
                list_lengh = len(todos)
            print(f"You have {list_lengh} item in your to do list.\n")
        case "exit":
            break

打印件(例如):

Enter add, show, edit, complete or exit to finish. : Add
Add an item: Apple
Enter add, show, edit, complete or exit to finish. : show
1-Apple

You have 1 item in your to do list.

Enter add, show, edit, complete or exit to finish. : exit

评论

1赞 Weverson Goncalves Muciarone 10/6/2023
这是伟大的安德烈。修复简单,效率高。非常感谢您的帮助。