"我的 Python 代码中的错误：Python 中的 elseif 错误"

Question

下面的代码不起作用

N = int(input())
print("Enter number",N)
 if(N%2!=0):
 print("Weird")
  elif (N%2==0) and (N>=2 and N<=5):
      print("Not Weird")
      elif(N%2==0) and (N>=6 and N<=20):
          print("Weird")
         else:
         print("Not Weird")

我收到以下错误：

File "solution.py", line 7
elif (N%2==0) and (N>=2 and N<=5):
   ^
SyntaxError: invalid syntax

Answer 1

问题在于缩进。

N = int(input())
print("Enter number",N)
if(N%2!=0):
   print("Weird")
elif (N%2==0) and (N>=2 and N<=5):
   print("Not Weird")
elif(N%2==0) and (N>=6 and N<=20):
   print("Weird")
else:
   print("Not Weird")

Answer 2

根据您的缩进，您不能在没有if情况下启动elif ，您应该始终启动if然后elif 。 您可以检查条件如下：

print("Enter number")
N = int(input())
if(N%2!=0):
    print("Weird")
elif (N%2==0) and (N>=2 and N<=5):
    print("Not Weird")
elif(N%2==0) and (N>=6 and N<=20):
    print("Weird")
else:
     print("Not Weird")

对于您的编码约定，您可以随时查看PEP 8 -- Python 代码风格指南

Answer 3

我的代码不起作用。 请帮忙！

if __name__ == '__main__':
    n = int(raw_input().strip())
if n%2! =0:
print("Weird")
else:
if n>=2 and n<=5:
print("Not Weird")
elif n>=6 and n<=20:
print("Weird")
elif n>20:
print("Not Weird")