如果..elif test与测试一起工作，那就太简单了

Question

我的任务是：

You have been asked to write a program that will give the name of a shape depending on the number of sides. The user can only enter numbers between 3 and 8, if they enter any other number then the program should tell them to enter a number between 3 and 8.

下面是我的答案：

#Sides and shapes
sides = int(input("How many sides on the shape are there? "))
if sides ==3:
    print ("Your shape is the triangle")
if sides ==4:
    print ("Your shape is the square")
if sides ==5:
    print ("Your shape is the pentagon")
if sides ==6:
    print ("Your shape is the hexagon")
if sides ==7:
    print ("Your shape is the heptagon")
if sides ==8:
    print ("Your shape is the octagon")
elif sides != range(3,9):
    print ("You should enter a number between 3 and 8")

在elif语句之后，我还需要以某种方式循环它，以便如果用户输入的不是3-8，那么它将继续要求用户输入3到8之间的数字。你知道吗

elif语句由于某种原因不起作用，并在F5中输出这个答案：

How many sides on the shape are there? 6
Your shape is the hexagon
You should enter a number between 3 and 8

Answer 1

range(x, y)输出介于x和y之间的数字范围。然后将一个整数（即边的值）与一个列表进行比较。显然它不等于列表，因为它是一个整数！你知道吗

Answer 2

range()产生一个不同类型的对象；integer != range()总是为真。你知道吗

使用<或<=和链接测试整数是否超出范围：

elif not (3 <= sides < 9):
    print ("You should enter a number between 3 and 8")

或者使用not in查看数字是否超出范围：

elif sides not in range(3, 9):
    print ("You should enter a number between 3 and 8")

或者只需对所有测试使用elif，但第一个测试除外，对最后一个分支使用else；仅当if..elif测试不匹配时才会选择它：

if sides ==3:
    print ("Your shape is the triangle")
elif sides ==4:
    print ("Your shape is the square")
elif sides ==5:
    print ("Your shape is the pentagon")
elif sides ==6:
    print ("Your shape is the hexagon")
elif sides ==7:
    print ("Your shape is the heptagon")
elif sides ==8:
    print ("Your shape is the octagon")
else:
    print ("You should enter a number between 3 and 8")

注意，现在只有一个if；从逻辑上讲，elif和else部分属于这个if语句。任何其他的if形成一个单独的、新的选择集，你的sides != range(3, 9)表达式总是正确的，这意味着elif测试在if slides == 8不正确的时候都是正确的。你知道吗

可以使用dictionary简化代码。它允许您将一个键与一个值相关联；将数字作为键，您可以简单地测试sides是否在字典中，或者返回一个默认值，如果不是：

shape_msg = "Your shape is the "
result = {
    3: shape_msg + "triangle",
    4: shape_msg + "square",
    5: shape_msg + "pentagon",
    6: shape_msg + "hexagon",
    7: shape_msg + "heptagon",
    8: shape_msg + "octagon",
}

sides = int(input("How many sides on the shape are there? "))
result = results.get(sides, "You should enter a number between 3 and 8")
print(result)

这里，^{} method返回给定键的值，如果键不存在，则返回默认值。你知道吗

如果需要继续循环，请基于以下内容测试是否存在密钥和分支：

while True:
    sides = int(input("How many sides on the shape are there? "))
    if sides in result:
        print(sides[result])
        break  # done, exit the loop
     print("You should enter a number between 3 and 8")

有关如何请求用户输入并处理错误输入的更多提示，请参见Asking the user for input until they give a valid response。你知道吗

Answer 3

您的elif只与前面的if关联，因此对于任何不是八角形的东西，它都会击中elif，然后正如其他人注意到的那样，您的比较并不是测试您是否在正确的范围内，而是将int与列表进行比较。你知道吗

可能您真的希望所有的if语句，除了第一个是elif，另一个作为elif的语句是整个块的else子句。你知道吗

更好的是，你可以有一个字典来映射消息的边数，如果字典中没有键，你可以打印错误消息。你知道吗

对于循环，如果你把整个东西放在while True中，你可以在每次成功之后break，而不是在无效的边数之后，这将导致循环重复，直到它们最终输入一个有效的数字。你知道吗