I want to create an if-condition, but I'm not sure how to.

[LeqendFire]

I want to create an if-condition, but I'm not sure how to do it... I would like, that if the else-condition is true, it prints my message. How can I do that?

That's my recent try:

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y =  'lol' if y == True:     print ('y ist wahr!') elif y == False:     print ('y ist nicht wahr!') else:     print ('y ist weder wahr noch nicht wahr')   if else == True:     print ('Ich habe es, denke ich, verstanden')

And that's the error I get:

Error:
if else == True:
         ^
SyntaxError: invalid syntax

I'm a starter, so it's probably easy to solve for you professionals

[Yoriz]

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something = False else_true = False   if something:     print('something is True') else:     print('something is False')     else_true = True   if else_true:     print('else_true is True')

[ndc85430]

It doesn't make sense for else to have a condition. Like in English, it means "in the other case". That's also why you can't have an else on its own.

[LeqendFire]

It doesn't make sense for else to have a condition. Like in English, it means "in the other case". That's also why you can't have an else on its own.

What do you mean with "you can't have an else on its own"? Could you explain that please?

[ndc85430]

As in, you can't write

1 2	else:   # something here

because again, else means "in the other case", so it has to follow an if or elif.

Of course, an if on its own makes sense. You might want to do something when the condition is True, but nothing if it is False.

[LeqendFire]

As in, you can't write

1 2	else:   # something here

because again, else means "in the other case", so it has to follow an if or elif.

Of course, an if on its own makes sense. You might want to do something when the condition is True, but nothing if it is False.

Alright, got it