Question

我已经定义了一个名为Start()的函数，我已经完成了这个函数，我想再次使用它，它将无法正常工作。它第一次工作，但如果我再次尝试使用此功能，它将无法正常工作。对不起，如果这个问题不明确。

import random

dice_numb = ["1", "2", "3", "4", "5", "6"]

def Start():
    print("Hello Welcome To Dice Rolling")
    roll = input("\nWould You Like To Roll? Y/N: ")
    while roll != "Y":
        if roll == "Y":
            print("Alright")
            randomindex = random.randint(0,len(dice_numb)-1) 

            print (dice_numb[randomindex])

            break
        if roll == "N":
            print("Okay,Cya")
            sys.exit()

    again = input("Would You Like To Roll Again? Y/N: ")

    while again != "Y":
        if again == "Y":
            print("Alright,Lets Do It!")
        if again -- "N":
            print("Okay,Cya.")
            sys.exit()
Start()

Answer 1

欢迎来到Python的世界:)。我在您的代码中看到了许多问题：

将这些更改与您的代码进行比较并阅读评论：

import random

def Start():                                          
    print("Hello Welcome To Dice Rolling")
    roll = input("\nWould You Like To Roll? Y/N: ")

    while roll == "Y":
            print("Alright")
            dice_no = random.randint(0, 6)              # or random.randint(0, len(dice_numb) -1)  
            print(dice_no)                             # No need to index list

            roll = input("\nWould You Like To Roll Again? Y/N: ")
    else: 
        print("Okay,Cya")

Start()

注意：不要使用大写字母启动函数名称，这种命名约定通常用于类而不是函数。

Answer 2

我不知道这是否有效;我也是python的新手。我想你忘了导入“sys”模块...

Answer 3

试试这样：

import random
import sys

dice_numb = ["1", "2", "3", "4", "5", "6"]

def Start():
    print("Hello Welcome To Dice Rolling")
    roll = input("\nWould You Like To Roll? Y/N: ")
    if roll.upper() == "Y":
        print("Alright")

        print random.choice(dice_numb)
    if roll.upper() == "N":
        print("Okay,Cya")
        sys.exit()
    while 1:
        again =  input("Would You Like To Roll Again? Y/N: ")

        if again.upper() == "Y":
            print("Alright,Lets Do It!")
        if again.upper() == "N":
            print("Okay,Cya.")
            break
Start()

当我尝试再次使用时，定义的功能无法打印？

3 个答案: