flag=False
if color1=="blue":
color2=input("Do you want to mix red or yellow with blue?")
color2=color2.lower()
while flag==False:
if color2=="red" or color2=="yellow":
flag = True
else:
color2=input("That is not a valid choice. Enter either red or yellow to mix\ with", color1)
if color2=="red":
print("The color you made is purple")
else:
print("The color you made is green")
flag=False
elif color1=="red":
color2=input("Do you want to mix blue or yellow with red?")
color2=color2.lower()
while flag==False:
if color2=="blue" or color2=="yellow":
flag = True
else:
color2=input("That is not a valid choice. Enter either blue or yellow to\ mix with", color1)
我遇到了一个问题,就是说“elif color1 ==”red“:”的行中间的一个elif语句。只是为了得到一些上下文,这是一个简单的程序,将混合三种基色。
答案 0 :(得分:3)
假设程序中的缩进与问题中显示的缩进相同,则flag=False
之前的行elif
需要缩进,否则解析器会认为你已经缩进当你实际上没有时,离开if
区块。