Question

我正在制作一个毕达哥拉斯计算器，但我一直在收到相同的信息＆＃34;语法无效＆＃34; 我是编码的新手，所以我只创建了我认为应该告诉用户缺少的一面，如果你能向我解释我做错了什么，我将非常感激。消息说＆＃34;语法无效＆＃34;如果x == 1＆＃39;

，则突出显示第一个冒号

# This should ask the user to pick an option

menu="Pythagoras Calculator:\n\
1. Find a hypotenuse?\n\
2. Find another side?\n\
9. Exit\n"


x = int(input(menu))

if x == 1:
    a = int(input("what is the other side?(cm)"))
    b = int(input("what is the last side?(cm)"))
    import math
    print("Hypotenuse=",math.sqrt((a**2)+(b**2)))

elif x == 2:
    c = int(input("what is the hypotenuse?(cm)"))
    d = int(input("what is the otherside?(cm)"))
    import math
    print("Other side=",math.sqrt((c**2)-(d**2)))

elif x == 9:
    print("Good Bye")

Answer 1

尝试将结束括号添加到x = int（输入（菜单）行

）

所以它应该是x = int（输入（菜单））

Answer 2

由于您不熟悉编码，我认为您可能想要更详细一些代码。

使用我机器上添加的右括号，您的代码可以正常工作。以下是一些可能有助于您从我的角度改进的建议：

在程序开始时加载数学模块 - 但是，如果你可以保持这种方式，并且在某些条件下会很好（比如你是否担心加载数学模块时的性能影响）不需要）。更多相关信息，如果您只想要平方根，则可以避免使用数学模块：((a**2)+(b**2))**0.5应该可以做到这一点。

非常重要：检查用户输入。你不能使负整数的平方根。所以你最好做

c = int(input("what is the hypotenuse?(cm)"))
d = int(input("what is the otherside?(cm)"))
assert d < c, "The otherside cannot be greater than the hypothenuse"
import math
print("Other side=",math.sqrt((c**2)-(d**2)))

始终确保用户的输入。这是计划中最大的问题来源之一。菜单也是如此。检查用户的输入是1,2或9.您可以设计一些内容以确保用户键入1,2或9，或者再次提示。

#With a printMenu function, you can print the menu again if you want to
def printMenu():
    menu="Pythagoras Calculator:\n\
          1. Find a hypotenuse?\n\
          2. Find another side?\n\
          9. Exit\n"
    print(menu)

#By setting x to None initially, you allow the following while loop
# to be executed at least once.
x = None
previouslyBad=False # This will be a flag in order to know if user got it wrong the first time
printMenu()

#With a loop, user will be asked for input again and again unless
# he uses the solutions you wanted him to use.
while x not in [1, 2, 9]:
    try:
        if not previouslyBad:
             x = int(input("Please choose a functionnality (e.g.: 1) "))
        else: 
             x = int(input("Please enter either 1, 2 or 9 "))
    #If you only use except without specifying ValueError,
    # user won't be able to use Ctrl+C to exit your program and will be
    # stuck in your menu. That is why you need to use two 
    # except statement.
    except ValueError:
        x = None
    except:
        raise
    previouslyBad = True #If the loop loops more than once, it should be aware of it

if x == 1:
    try:
        a = int(input("what is the other side?(cm)"))
        b = int(input("what is the last side?(cm)"))

    #Would you like to ensure entered values were int and not raise
    # error, you could use a similar strategy I implemented up with the
    # menu. 
    except ValueError:
        raise
    print("Hypotenuse=",((a**2)+(b**2))**0.5) # Using math.sqrt is fine too. But lazy loading indicated you would focus on performance. This might be slightly faster since it doesn't need any module loading.

elif x == 2:
    try:
        c = int(input("what is the hypotenuse?(cm)"))
        d = int(input("what is the otherside?(cm)"))
    except ValueError:
        raise
    #Assert here allows you to ensure user won't try squarerooting a
    # a negative value. 
    assert d <= c, "The otherside cannot be greater than the hypothenuse"
    print("Other side=",((c**2)-(d**2))**0.5)

elif x == 9:
    print("Good Bye")

我正在制作一个毕达哥拉斯计算器，但我一直在收到相同的信息＆＃34;语法无效＆＃34;

2 个答案: