Question

这里非常新，但我邀请3个人共进晚餐，我必须向他们发送一份来自名单的消息，邀请他们共进晚餐..我在开始时我已经

dinnerGuest = ['Steve Jobs', 'Tupac Shakur', 'Kobe Bryant']
message = """You have been cordially invited to an epic dinner.
Please RSVP as soon as possible. Thank you""" + dinnerGuest[0].title + "."
print(message)

我收到错误

Please RSVP as soon as possible. Thank you""" + dinnerGuest[0].title + "."
TypeError: Can't convert 'builtin_function_or_method' object to str implicitly

任何人都知道为什么？在此先感谢，我知道这是一个新手问题。我会到达那里。

Answer 1

dinnerGuest = ['Steve Jobs', 'Tupac Shakur', 'Kobe Bryant']
message = """You have been cordially invited to an epic dinner.
Please RSVP as soon as possible. Thank you""" + dinnerGuest[0].title() + "."
print(message)

title（）是一个函数 - 它需要一些东西并将其变成其他东西。

无论何时使用函数，总是将括号放在最后。

Answer 2

由于您想邀请列表中的所有访客，因此您必须使用循环遍历整个列表：

dinnerGuest = ['Steve Jobs', 'Tupac Shakur', 'Kobe Bryant']

message = """You have been cordially invited to an epic dinner. 
              Please RSVP as soon as possible. Thank you """

for i in dinnerGuest:
    print(message + i.title() + ".")

同样正如@Andrew Owen和@Francisco Couzo所提到的，你需要使用 title（）而不是 title ，因为它是一种方法，并且在它之后需要（）

处理python中的列表和消息

2 个答案: