解决 Python 中处理颜色输入时的常见错误

在 Python 中，当您尝试将字符串转换为整数时，可能会遇到 “invalid literal for int() with base” 错误。这通常发生在您尝试将颜色名称（例如 “红色” 或 “绿色”）转换为整数时。

lightcolor = int(input("Enter Red,Green,Yellow,White,Purple,Blue,Orange,Brown,or Black->"))

上面的代码尝试将用户输入的颜色名称转换为整数。但是，由于颜色名称不是整数，因此会引发 “invalid literal for int() with base” 错误。

解决方案

要解决这个问题，您可以使用以下方法：

使用 str 函数将颜色名称转换为字符串。
使用 raw_input() 函数而不是 input() 函数来获取用户输入。raw_input() 函数不会将用户输入转换为整数，因此不会引发错误。
使用 dict 来存储颜色名称和对应的整数。然后，您可以使用 get() 方法来获取对应于用户输入的颜色名称的整数。

代码例子

# 使用 str 函数将颜色名称转换为字符串
lightcolor = str(input("Enter Red,Green,Yellow,White,Purple,Blue,Orange,Brown,or Black->"))

# 使用 if-elif-else 语句来处理用户输入的颜色名称
if lightcolor == "Red":
    print("Red Light-Please stop!!")
elif lightcolor == "Green":
    print("Green Light-Please continue")
elif lightcolor == "Yellow":
    print("Yellow Light-speed up")
elif lightcolor == "White":
    print("White Light-its too bright")
elif lightcolor == "Purple":
    print("Purple Light-pretty")
elif lightcolor == "Blue":
    print("Blue Light-thats unusual")
elif lightcolor == "Orange":
    print("Orange Light-bright as the sun")
elif lightcolor == "Brown":
    print("Brown Light-like dirt")
elif lightcolor == "Black":
    print("Black Light-very dark")
else:
    print("Sorry no such color"),lightcolor

# 使用 raw_input() 函数来获取用户输入的颜色名称
lightcolor = raw_input("Enter Red,Green,Yellow,White,Purple,Blue,Orange,Brown,or Black->")

# 使用 if-elif-else 语句来处理用户输入的颜色名称
if lightcolor == "Red":
    print("Red Light-Please stop!!")
elif lightcolor == "Green":
    print("Green Light-Please continue")
elif lightcolor == "Yellow":
    print("Yellow Light-speed up")
elif lightcolor == "White":
    print("White Light-its too bright")
elif lightcolor == "Purple":
    print("Purple Light-pretty")
elif lightcolor == "Blue":
    print("Blue Light-thats unusual")
elif lightcolor == "Orange":
    print("Orange Light-bright as the sun")
elif lightcolor == "Brown":
    print("Brown Light-like dirt")
elif lightcolor == "Black":
    print("Black Light-very dark")
else:
    print("Sorry no such color"),lightcolor

# 使用 dict 来存储颜色名称和对应的整数
colors = {
    
    
    "Red": 1,
    "Green": 2,
    "Yellow": 3,
    "White": 4,
    "Purple": 5,
    "Blue": 6,
    "Orange": 7,
    "Brown": 8,
    "Black": 9
}

# 使用 get() 方法来获取对应于用户输入的颜色名称的整数
lightcolor = input("Enter Red,Green,Yellow,White,Purple,Blue,Orange,Brown,or Black->").lower()
color_number = colors.get(lightcolor, -1)

# 使用 if-elif-else 语句来处理用户输入的颜色名称
if color_number == 1:
    print("Red Light-Please stop!!")
elif color_number == 2:
    print("Green Light-Please continue")
elif color_number == 3:
    print("Yellow Light-speed up")
elif color_number == 4:
    print("White Light-its too bright")
elif color_number == 5:
    print("Purple Light-pretty")
elif color_number == 6:
    print("Blue Light-thats unusual")
elif color_number == 7:
    print("Orange Light-bright as the sun")
elif color_number == 8:
    print("Brown Light-like dirt")
elif color_number == 9:
    print("Black Light-very dark")
else:
    print("Sorry no such color"),lightcolor

上面的代码提供了三种不同的解决方案来解决 “invalid literal for int() with base” 错误。您可以根据自己的需要选择一种解决方案。

解决 Python 中处理颜色输入时的常见错误

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