Question

请通过以下代码为我提供帮助，

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")

我遇到以下错误：

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

请帮助我解决此错误，谢谢

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

基于缩进，如果没有RewriteEngine On RewriteCond %{HTTP_HOST} ^client-domain.in [NC] RewriteRule ^(.*)$ https://%{SERVER_NAME}%{REQUEST_URI}/$1 [L,R=301] RewriteCond %{HTTP_HOST} ^mydomain.in [NC] RewriteRule ^(.*)$ http://%{SERVER_NAME}%{REQUEST_URI}/$1 [L,R=301]，则无法启动elif，应始终先启动if，然后再启动if。您可以检查以下情况：

elif

对于您的编码约定，您始终可以查看PEP 8 -- Style Guide for Python Code

我的Python代码中的错误：Python中的elseif错误

2 个答案: