Python 3 语法错误无效语法
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Python 3 syntax error invalid syntax
提问by Bbrown
With this code I am trying to generate simple multiplication tables. The program should ask for input and multiple that number in a range up to 15 and the generate the multiplication table for the number. After the if_name_ == 'main': line I end up with a syntax error after the colon. I normally program in python 2, so python 3 is a bit new to me but I'm not sure what the difference is. Below I have listed the short but full code. Any help would be much appreciated.
使用此代码，我正在尝试生成简单的乘法表。程序应要求输入并将该数字乘以最多 15 的范围，并生成该数字的乘法表。在 if_name_ == ' main': 行之后，冒号后出现语法错误。我通常在 python 2 中编程，所以 python 3 对我来说有点新，但我不确定有什么区别。下面我列出了简短但完整的代码。任何帮助将非常感激。
'''Multiplication Table'''
def multi_table(a):
for i in range(1,16):
print(' {0} x {1} = {2} '.format(a, i, a*i))
if_name_ == '_main_':
a = input('Enter a number: ')
multi_table(float(a))
回答by Inconnu
if_name_ == '_main_':
a = input('Enter a number: ')
multi_table(float(a))
should be :
应该 ：
if __name__ == "__main__":
a = input('Enter a number: ')
multi_table(float(a))
Notice that both variable __name__and __main__has two underscores around them and that there must be a space between the if keyword and the start of the condition.
请注意， variable__name__和__main__周围有两个下划线，并且 if 关键字和条件的开头之间必须有一个空格。
回答by LMD
As @Maroun Maroun said right, it has to be if __name__ == "__main__". But you wont need it. Just write it at the bottom :
正如@Maroun Maroun 所说，它必须是if __name__ == "__main__". 但你不会需要它。直接写在底部：
'''Multiplication Table'''
def multi_table(a):
for i in range(1,16):
print(' {0} x {1} = {2} '.format(a, i, a*i))
a = input('Enter a number: ')
multi_table(float(a))
Should work, too.
也应该工作。
EDIT: In the official docs :
编辑：在官方文档中：
https://docs.python.org/3/library/main.html
https://docs.python.org/3/library/主要的.html
if __name__ == "__main__":
if __name__ == "__main__":
