Question

我做了一些代码，它会滚动两个模具并询问用户多少次。之后，它将询问用户是否想再次播放。出于某种原因，它应该在它应该的时候打破其中一个while循环。

以下是代码：

import random
again = True
rolls = 0
while again == True:
    rolls = float(raw_input("How many times would you like the dice to roll?"))
    while rolls >= 1:
        dice1 = random.randint(1,6)
        dice2 = random.randint(1,6)
        print dice1 , dice2
        rolls = rolls - 1
    again = raw_input("Would you like to play again?")
    if again == "Yes" or "Y" or "yes" or "y":
        again = True
    else:
        again = False

你们中的任何人都可以帮助我

Answer 1

问题在于这一行：

if again == "Yes" or "Y" or "yes" or "y":

评估为：

if (again == "Yes") or "Y" or "yes" or "y":

换句话说，它总是如此。你的意思是：

if again in ("Yes", "Y", "yes", "y"):

Answer 2

if again == "Yes" or "Y" or "yes" or "y":

这不符合你的想法。评估为：

if (again == "Yes") or ("Y") or ("yes") or ("y"):

非空字符串逻辑True，因此整个语句将始终为True。您需要将again与每个可能的值进行比较。

if again in ("Yes", "Y", "yes", "y"):
# or
if again == "Yes" or again == "Y" or again == "yes" or again == "y":

Answer 3

该行

if again == "Yes" or "Y" or "yes" or "y":

错误，将始终返回True。要修复它，请将其更改为

 if again == "Yes" or again == "Y" or again == "yes" or again == "y":

这种情况正在发生，因为非空字符串始终是真实的。由于您只是在第一个表达式中再次比较字符串，因此您只需检查字符串“Y”是否为空。由于该字符串是硬编码的，因此它始终为True

做起来也可能更整洁

#Uppercase the input and then check if user said yes
if again.upper() in ("YES","Y"):
    do_the_thing()

Answer 4

问题出在你的陈述中

if again == "yes" or "Y" or "yes" or "y"

python中的语句被评估为True，除非它们等于0或None，这是假的。

你必须这样做

if again == "Yes" or again == "Y" or again == "yes" etc.

这是因为or "Y" or "yes"总是评估为True。

Answer 5

下面的工作：

import random
again = True
rolls = 0
while again == True:
    rolls = float(raw_input("How many times would you like the dice to roll?"))
    while rolls >= 1:
        dice1 = random.randint(1,6)
        dice2 = random.randint(1,6)
        print dice1 , dice2
        rolls = rolls - 1
    again = raw_input("Would you like to play again?")
    # It correct and more short
    again = True if again in  ("Yes", "Y", "yes", "y") else False

没有打破while循环

5 个答案: