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。