On 11 October 2015 at 18:12, eetix letix <eetix.le...@gmail.com> wrote: > Hi, > > I'm sorry but the last version of Python (3.5.0) had a problem. I start and > I meet this problem : > >>>>a=5 >>>>if a>0: > . . . print("a is a positive.") > . . . if a<0: > ^ > SyntaxError: invalid syntax >>>> > > Normally this should work but problem comes to the fact that Python > considers "a" is a positive number and refuses to do the command >>>if a<0:
You're almost there, but your guess as to the reason for the error isn't quite right. It seems that the syntax in the Python shell is subtly different to code in a module. Because the second 'if' is a distinct statement from the first, the python shell seems to require a second return. However consider that perhaps using an 'else' or 'elif' instead of a completely separate 'if' And also consider that most people regard 0 to be positive, so you probably meant (a>=0). That means you can simplify to if a>=0: print("a is a positive") else: print("no need to evaluate a a second time") > And the command \n is doesn't working : > >>>> a="test\nto\nsee\nif\nit\nis\nworking" >>>> a > 'test\nto\nsee\nif\nit\nis\nworking' >>>> > > > Normally, \n should make that the text returns to the line but doesn't make > it. And if y do : > >>>> a="""test > . . . to > . . . see > . . . if > . . . it > . . . is > . . . working""" >>>>a > 'test\nto\nsee\nif\nit\nis\nworking' >>>> \n is an escape sequence rather than a command Have a look at what happens if you try print(a) > Thanks to fix this problems and good luck ;) > > > PS : I'm sorry for this really bad english but I'm french and I'm 14 Don't worry, it's certainly better than my French! -- Matt Wheeler http://funkyh.at -- https://mail.python.org/mailman/listinfo/python-list