Python 获取elif语句的无效语法错误。如何克服这个问题?
我正在尝试运行此代码,但出现了错误,并花费了所有时间进行调试。正在寻求帮助Python 获取elif语句的无效语法错误。如何克服这个问题?,python,python-3.x,if-statement,Python,Python 3.x,If Statement,我正在尝试运行此代码,但出现了错误,并花费了所有时间进行调试。正在寻求帮助 n = int(input()) if n % 2 == 1: print("Weird") elif(n%2==0) and 2<=n<=5: print("Not Weird") elif(n%2==0) and 6<=n<=20: print("Weird") else: print(" Not Weird") n=in
n = int(input())
if n % 2 == 1:
print("Weird")
elif(n%2==0) and 2<=n<=5:
print("Not Weird")
elif(n%2==0) and 6<=n<=20:
print("Weird")
else:
print(" Not Weird")
n=int(输入())
如果n%2==1:
印刷品(“怪异”)
elif(n%2==0)和2if/elif/else
语句都必须处于相同的缩进级别
if x:
# do stuff
elif y:
# do other stuff
elif z:
# do more stuff
else:
# do something else
你不应该缩进elif和else。它们应该是相同的缩进。此外,您不需要将条件语句封装在括号中。这项工作:
n = int(input())
if n%2 == 1:
print("Weird")
elif n%2 == 0:
print("Not Weird")
elif n%2 == 0 and 6 <= n <= 20:
print("Weird")
else:
print(" Not Weird")
n=int(输入())
如果n%2==1:
印刷品(“怪异”)
elif n%2==0:
打印(“不奇怪”)
elif n%2==0和6请添加代码。抱歉,由于缩进而花费了时间此处显示的缩进不正确