My "if" statements arent working

Question

OK so I'm not sure what im doing wrong here but my code is automatically choosing the first one in the "if" statement for exmaple if it is

if 1 == 2
print("Works")
elif 1 == 1
print("There we go")

it will automatically choose the first one even if the incorrect value is typed in. Please see my code below:

def troubleshoot ():
print("Now we will try and help you with your iPhone.")
print("")
time.sleep(1)
hardsoft = input(str("Is the issue a problem with hardware of software? Write n if you are not sure:  ") ) #ISSUE WITH IT SELECTING THE FIRST ONE NO MATER WHAT# 
if hardsoft == "Software" or "software" or "S" or "s" or "soft" or "Soft":
    software ()
elif hardsoft == "Hardware" or "hardware" or "Hard" or "hard" or "h" or "H":
    hardware ()
elif hardsoft == "Not sure" or "not" or "Not" or "NOT" or "not sure" or "n" or "N":
    notsure ()
else:
    print("Sorry, that command was not recognised")
    print("Please try again")
    troubleshoot ()

In your first code block, you're missing colons and indentation. Is that a miss copy-paste or does your code look exactly like that? — Celeo, Nov 21 '16 at 21:08
`if hardsoft == 'Software' or hardsoft == 'software' or hardsoft == 'S' ....` Or `if 'S' == hardsoft[:1].upper():` — 001, Nov 21 '16 at 21:10

score 1 · Accepted Answer · answered Nov 21 '16 at 21:14

1

In the first code block you are missing indentation and colons. It should be:

if 1 == 2:
    print("Works")
elif 1 == 1:
    print("There we go")

This way you also get the expected result.

For the second part: if hardsoft == "Software" or "software" or "S" or "s" or "soft" or "Soft": is not a valid condition - or at least it is not doing what you think it should do. Every string there gets converted to a boolean value, and afair any string that is not empty will be interpreted as true. Therefore a condition like if "Software" is always true. Correct condition would be:

if hardsoft == "Software" or hardsoft == "software" or hardsoft == "S" or hardsoft == "s" or hardsoft == "soft" or hardsoft == "Soft":
    ... and so on

answered Nov 21 '16 at 21:14

Striezel

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Thank you so much for your help – Leo Codron Nov 21 '16 at 21:17
My suggestion - `if hardsoft.lower() in ["software", "s", "soft"]:` – Peter Majko Nov 21 '16 at 21:18
Thats awesome, thanks again man – Leo Codron Nov 21 '16 at 21:21

score 0 · Answer 2 · answered Nov 21 '16 at 21:13

Your if statement contains many logical parts. Each part converted to boolean - so any non empty string converted to True. As I can suggest you wanted something like this:

def troubleshoot ():
print("Now we will try and help you with your iPhone.")
print("")
time.sleep(1)
hardsoft = input(str("Is the issue a problem with hardware of software? Write n if you are not sure:  ") ) #ISSUE WITH IT SELECTING THE FIRST ONE NO MATER WHAT# 
if hardsoft == "Software" or hardsoft == "software" or hardsoft == "S" or hardsoft == "s" or hardsoft == "soft" or hardsoft == "Soft":
    software ()
elif hardsoft == "Hardware" or hardsoft == "hardware" or hardsoft == "Hard" or hardsoft == "hard" or hardsoft == "h" or hardsoft == "H":
    hardware ()
elif hardsoft == "Not sure" or hardsoft == "not" or hardsoft == "Not" or hardsoft == "NOT" or hardsoft == "not sure" or hardsoft == "n" or hardsoft == "N":
    notsure ()
else:
    print("Sorry, that command was not recognised")
    print("Please try again")
    troubleshoot ()

Also you must be aware from such long ifs, and optimize this code.

Thank you so much for this, this is exactly what i was looking for. How would I optimize it? — Leo Codron, Nov 21 '16 at 21:16
You can predefine all your cases in arrays: sw = [ "Software", "software", "S","s","soft","Soft"], then just call sw.index[hardsoft] >= 0. — Oleg Imanilov, Nov 21 '16 at 21:21

My "if" statements arent working

2 Answers2