Elif invalid syntax python ошибка

I’m a newbie to Python and currently learning Control Flow commands like if, else, etc.

The if statement is working all fine, but when I write else or elif commands, the interpreter gives me a syntax error. I’m using Python 3.2.1 and the problem is arising in both its native interpreter and IDLE.

I’m following as it is given in the book ‘A Byte Of Python’ . As you can see, elif and else are giving Invalid Syntax.

>> number=23
>> guess = input('Enter a number : ')
>> if guess == number:
>>    print('Congratulations! You guessed it.')
>> elif guess < number:
   **( It is giving me 'Invalid Syntax')**
>> else:
   **( It is also giving me 'Invalid syntax')**

Why is this happening? I’m getting the problem in both IDLE and the interactive python. I hope the syntax is right.

answered Aug 11, 2011 at 12:08

elif and else must immediately follow the end of the if block, or Python will assume that the block has closed without them.

if 1:
    pass
             <--- this line must be indented at the same level as the `pass`
else:
    pass

In your code, the interpreter finishes the if block when the indentation, so the elif and the else aren’t associated with it. They are thus being understood as standalone statements, which doesn’t make sense.

In general, try to follow the style guidelines, which include removing excessive whitespace.

answered Aug 11, 2011 at 12:04

answered Aug 11, 2011 at 14:08

 if guess == number:
     print ("Good")
 elif guess == 2:
     print ("Bad")
 else:
     print ("Also bad")

Make sure you have your identation right. The syntax is ok.

Remember that by default the return value from the input will be a string and not an integer. You cannot compare strings with booleans like <, >, =>, <= (unless you are comparing the length). Therefore your code should look like this:

number = 23
guess = int(input('Enter a number: '))  # The var guess will be an integer

if guess == number:
    print('Congratulations! You guessed it.')

elif guess != number:
    print('Wrong Number')

answered Feb 15, 2017 at 2:42

Besides that your indention is wrong. The code wont work.
I know you are using python 3. something.
I am using python 2.7.3
the code that will actually work for what you trying accomplish is this.

number = str(23)
guess =  input('Enter a number: ')
if guess == number:
   print('Congratulations! You guessed it.')
elif guess < number:
     print('Wrong Number')
elif guess < number:
     print("Wrong Number')

The only difference I would tell python that number is a string of character for the code to work. If not is going to think is a Integer. When somebody runs the code they are inputing a string not an integer. There are many ways of changing this code but this is the easy solution I wanted to provide there is another way that I cant think of without making the 23 into a string. Or you could of «23» put quotations or you could of use int() function in the input. that would transform anything they input into and integer a number.

answered Apr 12, 2013 at 21:39

answered May 14, 2017 at 3:29

indentation is important in Python. Your if else statement should be within triple arrow (>>>), In Mac python IDLE version 3.7.4 elif statement doesn’t comes with correct indentation when you go on next line you have to shift left to avoid syntax error.

answered Jul 29, 2019 at 2:07

Отступы при таких операторах должны быть одинаковыми:

if cond:
    ...
elif cond:
    ...
else:
    ...

04 июн. 2023, в 01:35

1500 руб./за проект

04 июн. 2023, в 01:25

40000 руб./за проект

03 июн. 2023, в 23:42

1500 руб./за проект

i’m having an invalid syntax on my elif statements why heres the code:

while correct_turns != len(word):
        if correct_turns == len(word):
            finish()
        guess = input("Please guess a letter from above: ")
        elif guess in word and not LettersUnused:
            used_letter(guess,display,LettersUnused)
        elif guess in word and LettersUnused:
            correct(guess,display,correct_turns,LettersUnused)
        elif guess in LettersUnused and not word:
            incorrect(guess,display,LettersUnused)
        elif guess not in LettersUnused:
            used_letter(guess,display,LettersUnused)
        elif len(guess)>1:
            guess_word(guess)

answered Apr 21, 2014 at 14:05