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Question

我正在为学校做一个小项目，我们需要将难度设置为1到3，但是当有人输入错误的数字时，他们会收到一行提示，请在1到3之间进行选择，但是问题应该自己重复，现在当您输入错误的数字时，代码才结束。

difficulty = int(input("Difficulty: "))

while 0 > difficulty > 4:
    print("This is not a valid difficulty, please choose 1, 2 or 3")
    difficulty = int(input("Difficulty: "))

if 0 < difficulty < 4:
    print("The playing board was created with difficulty: " + str(difficulty))

Answer 1

while循环0 > difficulty > 4永远不会执行，因为该条件总是评估为False，因为0 > 4是False，因此我将while循环重构为{{1 }}会检查难度是否小于0或大于4，也正如@deceze指出的那样，while difficulty > 4 or difficulty < 0:是不需要的，因为条件只有在我们确保难度在0到4之间时才会出现，不包括0和4

所以答案变为

if

输出将类似于

difficulty = int(input("Difficulty: "))

#Check if difficulty is less than 0, or greater than 4
while difficulty < 0 or difficulty > 4:
    print("This is not a valid difficulty, please choose 1, 2 or 3")
    difficulty = int(input("Difficulty: "))

print("The playing board was created with difficulty: " + str(difficulty))

编写while循环的另一种方法是，我们需要确保如果输入小于0或大于4，我们希望继续运行循环，实际上可以通过Difficulty: -1 This is not a valid difficulty, please choose 1, 2 or 3 Difficulty: 5 This is not a valid difficulty, please choose 1, 2 or 3 Difficulty: 2 The playing board was created with difficulty: 2 < / p>

然后答案将更改为

while not 0 < difficulty < 4:

Answer 2

尝试这样：

difficulty = int(input("Enter input :"))
while difficulty<1 or difficulty>3:
  difficulty = int(input("Enter input between 1 and 3 :"))
print("correct input:",difficulty)

Answer 3

“难度低于0而难度大于4的情况” 永远不可能成立，因为没有数字同时小于0和大于4。格式化条件的最易读方法是使用range：

difficulty = int(input("Difficulty: "))

while difficulty not in range(1, 4):
    print("This is not a valid difficulty, please choose 1, 2 or 3")
    difficulty = int(input("Difficulty: "))

print("The playing board was created with difficulty: " + str(difficulty))

您也可以省略if，因为循环已经确保该值在有效范围内；无需再次检查。

Answer 4

尝试一点递归！

def getDiff():
    difficulty = int(input("Difficulty: "))

    if 0 < difficulty < 4:
        print("The playing board was created with difficulty: " + str(difficulty))
        return 
    else:
        print("This is not a valid difficulty, please choose 1, 2 or 3")
        getDiff()

getDiff()

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返回难度

def getDiff():
    difficulty = int(input("Difficulty: "))

    if 0 < difficulty < 4:
        print("The playing board was created with difficulty: " + str(difficulty))
        return difficulty 
    else:
        print("This is not a valid difficulty, please choose 1, 2 or 3")
        getDiff()

difficulty = getDiff()

重复一个问题，直到一个好的答案来

4 个答案:

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